Particle Duality
Questions for Short Answer
1. Can we find the mass of a photon by the definition p =mv?
ANSWER: p =mv is the momentum of a particle according to classical physics. In classical physics, light/electromagnetic radiation is a wave propagation, and it does not have a mass. The photon name itself is the result of quantum physics. According to this theory,
The momentum of a photon is not due to its mass (mass is assumed to be zero) but it is without its mass. The energy of a photon E is given as,
E² =(mc²)²+(pc)²
Since mass id zero, E =pc.
→p =E/c. Clearly, this momentum is not due to its mass. So we can not find the mass of a photon by p =mv.
2. Is it always true that for two sources of equal intensity, the number of photons emitted in a given time is equal?
ANSWER: The intensity of light is the energy passing normally to a unit area in unit time. Two light sources may have the same intensity even if they have different wavelengths. In this case, the number of photons emitted in a given time will be different because the energy of a photon depends on its wavelength. Only if the wavelengths are the same, the number will be the same.
3. What is the speed of a photon with respect to another photon if (a) the two photons are going in the same direction and (b) they are going in opposite directions?
ANSWER: A photon always travels at a speed equal to the speed of light c. This is true for any frame of reference used to observe the photon. So in both the given conditions, if we chose one photon as a frame of reference to get the relative speed of the other one, its speed will always be equal to c.
4. Can a photon be deflected by an electric field? By a magnetic field?
ANSWER: Since a photon is electrically or magnetically neutral, it will neither be deflected by an electric field nor by a magnetic field.
5. A hot body is kept in a closed room maintained at a lower temperature. Is the number of photons in the room increasing?
ANSWER: Photons are not like objects that can be kept in a closed room. The hot body will radiate a part of losing heat in the form of electromagnetic radiation that essentially is through photons. These photons will strike room walls and will be absorbed there. Soon there will be a balance there and if the temperature of the hot body is maintained, the number of photons emitted and absorbed will be equal. Thus at any instant, the number of photons will be constant.
6. Should the energy of a photon be called its kinetic energy or its internal energy?
ANSWER: A photon is itself a quantum of energy. It has no mass and always a speed equal to c. The energy associated with a photon is due to its frequency and it is equal to h𝛎. According to classical physics, the kinetic energy of an object depends on its mass and velocity. Photons have zero mass and constant velocity c but different photons of different frequencies have different energies. On this basis, its energy can not be called kinetic energy.
A photon has no internal structure hence it has no internal energy.
So a photon is itself a packet of energy that depends on its frequency.
7. In an experiment on the photoelectric effect, a photon is incident on an electron from one direction and the photoelectron is emitted almost in opposite direction. Does this violet conservation of momentum?
ANSWER: We must not forget that the colliding electron is not an independent electron but it is a part of an atom. When considering the conservation of momentum, we should count the incident photon and the atom as a system. Total energy and total momentum are conserved in such collisions. Thus there is no violation of the conservation of momentum.
8. It is found that yellow light does not eject photoelectrons from a metal. Is it advisable to try with orange light? With a green light?
ANSWER: Since yellow light does not eject photoelectrons from metal, it means that the frequency of yellow light is less than the threshold frequency of the metal. We need a greater frequency of light to get photoelectrons. Since the frequency of the orange light is less than the yellow light it is not advisable to try this one. The frequency of the green light is more than that yellow light, so there is a chance to get photoelectrons. Hence it is advisable to try the green light.
9. It is found that photosynthesis starts in certain plants when exposed to sunlight but it does not start if the plant is exposed only to infrared light. Explain.
ANSWER: Photosynthesis in plants starts with the help of a substance called chlorophyll that is present in green leaves. It needs a photon of sufficient energy to be absorbed by its electron and get excited to be released. This breaks the water molecule and with the presence of carbon dioxide photosynthesis starts. The sunlight has photons of sufficient energy to excite the chlorophyll molecules and begin the process but infrared light has a lower frequency than sunlight and its photons have lower energies. These photons are unable to excite the chlorophyll molecules and photosynthesis does not start.
10. The threshold wavelength of a metal is λₒ. Light of wavelength slightly less than λₒ is incident on an insulated plate made of this metal. It is found that photoelectrons are emitted for some time and after that, the emission stops. Explain.
ANSWER: The energy of light of wavelength slightly less than the threshold wavelength λₒ will have energy more than the work function. So the photoelectrons start to emit from the surface of the insulated metal plate after getting energy from the incident photons. These are the detected photoelectrons. Since the plate is insulated the deficiency of the electrons can not be filled. Thus a net positive charge will be there on the surface of the plate. So further electrons can not get out of the surface due to the electrostatic attraction and the emission of photoelectrons stops.
11. Is p =E/c valid for electrons?
ANSWER: No. The photons have constant speed 'c', and the energy of each photon is E =hc/λ. Since the momentum of a photon is p =h/λ, hence E =pc
→p =E/c.
But the electrons have speeds much less than 'c', so E ≠ hc/λ. So p =E/c is not valid for electrons.
12. Consider the de Broglie wavelength of an electron and a proton. Which wavelength is smaller if the two particles have (a) the same speed (b) the same momentum (c) the same energy?
ANSWER: De Broglie's wavelength is,
λ =h/p,
where h is Plank's constant and p is the momentum of the particle.
(a) Since a proton is much heavier than an electron, the proton will have greater momentum than the electron for the same speed. Clearly, the proton will have a smaller wavelength.
(b) When both have the same momentum p, wavelength λ =h/p will be the same for both.
(c) For two particles having mass m and m', speed v and v', if both have equal kinetic energy then,
mv² =m'v'²
Multiplying both sides by mm', we get
mm'mv² =mm'm'v'²
→m²v²m' =m'²v'²m
→p²m' =p'²m
Where p is the momentum of the particle having m and p' for particle m'.
→(p'/p)² =m'/m
→p'/p =√(m'/m)
If m' > m then p' > p.
Since the mass of the proton > the mass of the electron, the momentum of the proton > the momentum of the electron for equal energy.
We have, λ =h/p, hence the wavelength of the proton will be less than the wavelength of the electron for equal energy.
13. If an electron has a wavelength, does it also have a color?
ANSWER: The color we see is due to those electromagnetic waves that have wavelengths within a certain range that strike our retina and excite it. The wavelength associated with an electron is not the wavelength of an electromagnetic wave but it is that of de Broglie's waves. Hence it has no color.
OBJECTIVE - I
1. Plank Constant has the same dimensions as
(a) force x time
(b) force x distance
(c) force x speed
(d) force x distance x time
ANSWER: (d).
Explanation: Plank Constant,
h = E/𝝂
Dimensions of energy E are the same as
Force x distance.
Unit of frequency 𝝂 is the number per unit of time. So the dimensions of h are the same as
Force x distance x time.
Hence option (d) is correct.
2. Two photons having
(a) equal wavelengths have equal linear momenta
(b) equal energies have equal linear momenta
(c) equal frequencies have equal linear momenta
(d) equal linear momenta have equal wavelengths.
ANSWER: (d).
Explanation: Linear momentum is a vector, that is it has a specific direction. While in the first three options, the magnitudes of linear momenta may be the same but the directions may be different. If their directions are different, they can not be said the same. Option (d) is correct.
3. Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,
(a) both p and E increase
(b) p increases and E decreases
(c) p decreases and E increases
(d) both p and E decrease.
ANSWER: (a).
Explanation: p = E/c and E =hc/λ.
So, p =h/λ.
Since h and c are constants, both p and E are inversely proportional to λ.
Thus if λ is decreased, both p and E increase. Option (a) is correct.
4. Let nᵣ and nᵦ be respectively the number of photons emitted by a red bulb and a blue bulb of equal power in a given time.
(a) nᵣ = nᵦ
(b) nᵣ < nᵦ
(c) nᵣ > nᵦ
(d) The information is insufficient to get a relation between nᵣ and nᵦ.
ANSWER: (c).
Explanation: Both the bulbs emit equal energy say 'E*' per second. So,
nᵣEᵣ = nᵦEᵦ =E* -------- (a)
where Eᵣ and Eᵦ are energies of photons emitted by red and blue bulbs respectively.
Since energy of a photon = hc/λ, from (a),
nᵣhc/λᵣ = nᵦhc/λᵦ
→nᵣ/nᵦ =λᵣ/λᵦ
We know that the wavelength of red light is more than the wavelength of blue light. Hence the number of photons emitted by the red bulb will be more than the number of photons emitted by the blue bulb. nᵣ > nᵦ. Option (c) is correct.
5. The equation E = pc is valid
(a) for an electron as well as for a photon
(b) for an electron but not for a photon
(c) for a photon but not for an electron
(d) neither for an electron nor for a photon.
ANSWER: (c).
Explanation: Equation E = pc is valid for a particle that has zero rest mass. Since a photon has zero rest mass but an electron has a rest mass. So it is only correct for a photon. Option (c) is correct.
6. The work function of a metal is h𝝂ₒ. Light of frequency 𝝂 falls on this metal. The photoelectric effect will take place only if
(a) 𝝂 ≥ 𝝂ₒ
(b) 𝝂 > 2𝝂ₒ
(c) 𝝂 < 𝝂ₒ
(d) 𝝂 < 𝝂ₒ/2.
ANSWER: (a).
Explanation: Work function is the minimum energy required to detach an electron from the surface of a metal. The energy of an incident photon is h𝝂. This energy should be more than or equal to the work function for the emission of a photoelectron. So,
h𝝂 ≥ h𝝂ₒ
→ 𝝂 ≥ 𝝂ₒ
Option (a) is correct.
7. Light of wavelength λ falls on a metal having work function hc/λₒ. The photoelectric effect will take place only if
(a) λ ≥ λₒ
(b) λ ≥ 2λₒ
(c) λ ≤ λₒ
(d) λ < λₒ/2.
ANSWER: (c).
Explanation: The energy of a photon having wavelength λ is
=hc/λ.
The photoelectric effect will take place only if this energy is greater than or equal to the work function, So,
hc/λ ≥ hc/λₒ
→1/λ ≥ 1/λₒ
→λ ≤ λₒ
Option (c) is correct.
8. When stopping potential is applied in an experiment on the photoelectric effect, no photocurrent is observed. This means that
(a) the emission of photoelectrons is stopped.
(b) the photoelectrons are emitted but are reabsorbed by the emitter metal
(c) the photoelectrons are accumulated near the collector plate
(d) the photoelectrons are dispersed from the sides of the apparatus.
ANSWER: (b).
Explanation: When photons of light having energies more than the work function of a metal are incident upon it, photoelectrons are emitted. Out of the total energy of a photon, energy equal to the work function is used just to take the electron out of the metal, and the rest of the energy is used as the kinetic energy of the electron. Due to this kinetic energy, the electron moves to the other end of the apparatus where it is collected and the photocurrent starts.
When stopping potential is applied, the collecting plate is made negative compared to the emitter plate. This repels the photoelectrons towards the emitter plate and some of them are reabsorbed there. Thus the photocurrent stops. Option (b) is correct.
9. If the frequency of light in a photoelectric experiment is doubled, the stopping potential will
(a) be doubled
(b) be halved
(c) become more than double
(d) become less than double.
ANSWER: (c).
Explanation: If V is stopping potential and φ the work function, then
eV = h𝝂 -φ
→V = (h𝝂-φ)/e
If the frequency of the light is doubled to 2𝝂, let the stopping potential be equal to V'. So,
V' = (2h𝝂-φ)/e
={(2h𝝂-2φ) +φ}/e
=2(h𝝂-φ)/e +φ/e
=2V +φ/e
So V' is more than 2V. Option (c) is correct.
10. The frequency and intensity of a light source are both doubled. Consider the following statements.
(A) The saturation photocurrent remains almost the same.
(B) The maximum kinetic energy of the photoelectrons is doubled.
(a) Both A and B are true.
(b) A is true but B is false.
(c) A is false but B is true.
(d) Both A and b are false.
ANSWER: (b).
Explanation: Keeping frequency constant, if the intensity of light is increased then the number of photons increases resulting in more photoelectrons. This increases the photocurrent. But when we double the frequency of light, the intensity is also doubled (since intensity is the amount of light energy falling per unit area, doubling the frequency also doubles the energy of photons). It means the number of photons does not increase. Each photon now carries double energy and the intensity of light is the energy falling on a unit area in unit time. Since the number of photons is not increasing, the number of photoelectrons is also not increasing. The saturation photocurrent will be almost the same. Statement A is correct.
The increased energies of the photons will be used to increase the maximum kinetic energy of the emitted photoelectrons.
Kₘₐᵪ =h𝝂 -φ, where φ is the work function.
If the frequency is doubled as in this case,
K'ₘₐᵪ =2h𝝂 -φ
Clearly, the last one is not double the original maximum kinetic energy. Statement B is incorrect.
Option (b) is correct.
11. A point source of light is used in a photoelectric effect. If the source is removed farther from the emitting metal, the stopping potential
(a) will increase
(b) will decrease
(c) will remain constant
(d) will either increase or decrease.
ANSWER: (c).
Explanation: When the point source of light is moved farther from the metal, the intensity of light near the metal will decrease but the frequency remains the same. The stopping potential depends on the frequency of light, not on the intensity. So the stopping potential will remain constant in this case. Option (c) is correct.
12. A point source causes a photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent as a function of the distance between the source and the metal?  |
| The figure for Q-12 |
ANSWER: (d).
Explanation: If the point source emits energy E per second then the intensity of light at distance r is
I =E/4πr².
Since the saturation photocurrent, 'i' is directly proportional to intensity, so
i ∝ 1/r².
This relationship is best represented by curve d in the figure.
Option (d) is correct.
13. Nonmonochromatic light is used in an experiment on the photoelectric effect. The stopping potential
(a) is related to the mean wavelength
(b) is related to the longest wavelength
(c) is related to the shortest wavelength
(d) is not related to the wavelength.
ANSWER: (c).
Explanation: Stopping potential depends on the wavelength of the light. The smaller the wavelength, the greater the energy of the photons, and hence more the stopping potential. To stop the photocurrent in this case, photoelectrons resulting from the shortest wavelength of photons should be stopped.
Option (c) is correct.
14. A proton and an electron are accelerated by the same potential difference. Let λₑ and λₚ denote the de Broglie wavelengths of the electron and proton respectively.
(a) λₑ = λₚ
(b) λₑ < λₚ
(c) λₑ > λₚ
(d) The relation between λₑ and λₚ depends on the accelerating potential difference.
ANSWER: (c).
Explanation: When a proton and an electron are accelerated through the same potential difference, they gain the same kinetic energies because of the same magnitude of the charge. If v is the speed gained by the electron through a potential difference of V then,
½mₑv² =eV
→(mₑv)² =2mₑeV
→pₑ =√(2mₑeV)
Where pₑ is the momentum of the electron.
Similarly, the momentum of the proton
pₚ =√(2mₚeV)
Since the mass of a proton is more than the mass of an electron, so
pₚ > pₑ
Now de Broglie's wavelength is given as,
λ = h/p,
Electron's wavelength λₑ =h/pₑ
Proton's wavelength λₚ =h/pₚ
Clearly, λₑ > λₚ.
Option (c) is correct.
OBJECTIVE - II
1. When the intensity of a light source is increased,
(a) the number of photons emitted by the source in unit time increases
(b) the total energy of the photons emitted per unit of time increases
(c) more energetic photons are emitted
(d) faster photons are emitted.
ANSWER: (a), (b).
Explanation: When the intensity of light is increased, it means the energy emitted by the light source per unit of time is increased without changing the wavelength of emitted light. This increase in energy is achieved by increasing the number of photons emitted in a unit of time. Since the wavelength is the same, the energy of an individual photon remains the same but the total energy of photons emitted per unit of time increases due to an increase in the number of photons. So options (a) and (b) are correct but (c) are incorrect.
Since the speed of photons is always the same whatever may be the wavelength, there is no question of faster photons. Option (d) is not correct.
2. The photoelectric effect supports the quantum nature of light because
(a) there is a minimum frequency below which no photoelectrons are emitted
(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not its intensity
(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately
(d) the electric charge of the photoelectrons is quantized.
ANSWER: (a), (b), (c).
Explanation: If the quantum nature of light were not true, the maximum kinetic energy should be dependent on the intensity because the intensity is the amount of energy falling on a unit area of a cross-section in a unit of time. Also, photoelectrons should be emitted with any wavelength by just changing the intensity. But it is not found true. The energy of a photon is dependent on its frequency. A photoelectron is emitted from a particular surface only if a minimum amount of energy is gained by it through the photon. So for this photo effect to take place a minimum frequency of the photon is required. So even if the surface is faintly illuminated with this frequency or a higher frequency of photons, photoelectrons leave the surface immediately. If the higher frequency of the photons is incident on the surface than the required frequency, maximum kinetic energy is increased.
Thus the energy of photons is quantized and related to the frequency. The options (a), (b) and (c) are correct.
The electric charge of a photoelectron is constant. Option (d) is not correct.
3. A photon of energy h𝜈 is absorbed by a free electron of a metal having work function φ <h𝜈.
(a) the electron is sure to come out.
(b) the electron is sure to come out with kinetic energy h𝜈-φ
(c) either the electron does not come out or it comes out with a kinetic energy h𝜈-φ.
(d) it may come out with kinetic energy less than h𝜈-φ.
ANSWER: (d).
Explanation: When the free electron of a metal absorbs a photon of energy h𝜈, it gets a kinetic energy =h𝜈-φ, but it may lose some of the absorbed kinetic energy due to collision with adjacent molecules. So it may come out with kinetic energy =h𝜈 -φ. Option (d) is correct.
4. If the wavelength of light in an experiment on the photoelectric effect is doubled,
(a) the photoelectric emission will not take place
(b) the photoelectric emission may or may not take place
(c) the stopping potential will increase
(d) the stopping potential will decrease.
ANSWER: (b), (d).
Explanation: With a doubled wavelength, the energy of photons will be halved because E =hc/λ. Only if the halved energy is more than the work function, photoelectric emission may take place. So option (b) is correct.
The maximum kinetic energy of electrons (hc/𝜆 -φ) will get decreased hence less stopping potential will be needed. Option (d) is correct.
5. The photocurrent in an experiment on the photoelectric effect increases if
(a) the intensity of the source is increased
(b) the exposure time is increased
(c) the intensity of the source is decreased
(d) the exposure time is decreased.
ANSWER: (a).
Explanation: Photocurrent will increase if the number of photoelectrons increases. The number of photoelectrons will increase if the number of photons increases. The number of photons will increase if the intensity of the source is increased. Photocurrent does not depend on exposure time. Option (a) is correct.
6. The collector plate in an experiment on the photoelectric effect is kept vertically above the emitter plate. The light source is put on and a saturation photocurrent is recorded. An electric field is switched on which has a vertically downward direction.
(a) the photocurrent will increase
(b) the kinetic energy of electrons will increase
(c) the stopping potential will decrease
(d) the threshold wavelength will increase.
ANSWER: (b).
Explanation: Since the direction of the electric field is downwards, it will apply an upward force on the emitted photoelectrons. The kinetic energy of photoelectrons will increase. Option (b) is correct. The current depends on the number of electrons emitted per unit of time which is not changing here. So photocurrent will not increase. Option (a) is not correct. Since kinetic energy increases, the stopping potential will increase, not decrease. Option (c) is not correct.
Threshold wavelength is a property of the emitter surface, so it will not change with the application of an electric field. Option (d) is not correct.
7. In which of the following situations does the heavier of the two particles have a smaller de Broglie wavelength? The two particles
(a) move with the same speed
(b) move with the same linear momentum
(c) move with the same kinetic energy
(d) have fallen through the same height.
ANSWER: (a), (c), (d).
Explanation: De Broglie's wavelength is,
𝜆 =h/p, where p is the linear momentum.
When both particles have the same speed, the linear momentum of a heavier particle will be more. So de Broglie wavelength of the heavier particle will be smaller. Option (a) is correct.
When both particles have the same linear momentum p, they both have the same de Broglie wavelength,
𝜆 =h/p. Option (b) is incorrect.
Kinetic energy, K =½mv²
→K =½(mv)²/m =p²/2m
→p =√(2mK)
So the heavier particle will have greater linear momentum and hence the shorter de Broglie wavelength 𝜆. Option (c) is correct.
When two particles fall through the same height their speeds are the same. As in option (a), the heavier particles will have greater linear momentum p and hence the smaller wavelength 𝜆. Option (d) is also correct.
EXERCISES
1. Visible light has wavelengths in the range of 400 nm to 780 nm. Calculate the range of energy of the photons of visible light.
ANSWER: The energy of a photon,
E =hc/λ.
For λ =780 nm =780x10⁻⁹ m
=7.80x10⁻⁷ m
c =3x10⁸ m/s, h =6.63x⁻³⁴ J-s.
So, E =6.63x10⁻³⁴*3x10⁸/7.8x10⁻⁷ J
=2.55x10⁻¹⁹ J.
For λ =400 nm
=4x10⁻⁷ m
E =6.63x10⁻³⁴*3x10⁸/4x10⁻⁷ J
≈5.0x10⁻¹⁹ J
So the range of energy of photons of visible light is from 2.55x10⁻¹⁹ J to 5.0x10⁻¹⁹ J.
2. Calculate the momentum of a photon of light of wavelength 500 nm.
ANSWER: The momentum of a photon of wavelength λ is given as,
p =h/λ
For λ =500 nm =5x10⁻⁷ m
The momentum of the photon,
p =6.63x10⁻³⁴/5x10⁻⁷ kg-m/s
=1.33x10⁻²⁷ kg-m/s.
3. An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 700 nm. Find the net energy absorbed by the atom in the process.
ANSWER: The energy of a photon of wavelength 500 nm i.e. 5x10⁻⁷ m is,
E =hc/λ
=6.63x10⁻³⁴*3x10⁸/5x10⁻⁷ J
=3.9x10⁻¹⁹ J.
The energy of a photon of wavelength λ =700 nm =7x10⁻⁷ m.
E' =6.63x10⁻³⁴*3x10⁸/7x10⁻⁷ J
=2.8x10⁻¹⁹ J.
Hence the net energy absorbed by the atom in this process =E -E'
=3.9x10⁻¹⁹ -2.8x10⁻¹⁹ J
=1.1x10⁻¹⁹ J.
4. Calculate the number of photons emitted per second by a 10 W sodium vapor lamp. Assume that 60% of the consumed energy is converted into light. The wavelength of Sodium light =590 nm.
ANSWER: The energy of a photon of sodium light of wavelength λ =590 nm =5.9x10⁻⁷ m is
E =hc/λ
=6.63x10⁻³⁴*3x10⁸/5.9x10⁻⁷ J
=3.37x10⁻¹⁹ J.
10% of 10 W i.e. 6 W of energy is converted into light. So the energy of 6 J/s is being emitted. Hence the number of photons emitted per second is
=6/3.37x10⁻¹⁹
=1.78x10¹⁹.
5. When the light is directly overhead, the surface of the earth receives 1.4x10³ W/m² of sunlight. Assume that the light is monochromatic with an average wavelength of 500 nm and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is 1.5x10¹¹ m. (a) Calculate the number of photons falling per second on each square meter of the earth's surface directly below the sun. (b) How many photons are there in each cubic meter near the earth's surface at any instant? (c) How many photons does the sun emit per second?
ANSWER: (a) The light energy falling per m² of earth's surface in a second,
E' =1.4x10³ J
The average wavelength of the falling light,
λ =500 nm =5.0x10⁻⁷ m.
The energy of a photon of this light,
E =hc/λ
=6.63x10⁻³⁴*3x10⁸/5.0x10⁻⁷ J
=3.98x10⁻¹⁹ J
The number of photons falling per second on each square meter of the earth's surface directly below the sun,
=E'/E
=1.4x10³/3.98x10⁻¹⁹
=3.5x10²¹.
(b) In a second, a photon travels 3x10⁸ m and 3.5x10²¹ number of photons fall on an m² of area. Thus there are 3.5x10²¹ photons in a volume of,
3.0x10⁸ m*1 m² =3.0x10⁸ m³.
So the number of photons in 1 m³ near the earth's surface,
=3.5x10²¹/3.0x10⁸
≈1.2x10¹³.
(c) The distance between the sun and the earth,
R =1.5x10¹¹ m.
Keeping the sun at the center, the area of a sphere of radius R is =4πR². The total number of photons P, emitted per second by the sun is falling on this area 4πR². The number of photons falling per second on each square meter of the area is 3.5x10²¹. Hence the number of photons emitted per second by the sun is,
=3.5x10²¹*4π*(1.5x10¹¹)²
=99x10⁴³
=9.9x10⁴⁴.
6. A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle of incidence is 60° and the number of photons striking the mirror per second is 1.0x10¹⁹. Calculate the force exerted by the light beam on the mirror.
ANSWER: Wavelength of the monochromatic light,
λ =663 nm =6.63x10⁻⁷ m
Linear momentum of a photon
p =h/λ
=6.63x10⁻³⁴/6.63x10⁻⁷ kg-m/s
=1.0x10⁻²⁷ kg-m/s
The number of photons striking per second =1.0x10¹⁹
Hence the total momentum of the photons falling per second on the mirror,
p' =1.0x10¹⁹*1.0x10⁻²⁷ kg-m/s
=1.0x10⁻⁸ kg-m/s
 |
| Diagram for Q-6 |
The direction of this momentum is at 60° from the normal to the mirror. Since the angle of reflection is equal to the angle of incidence, the reflected photons will also have the same magnitude of momentum but its direction will be 60° from the normal and going away from the mirror. If we resolve the momenta of the incoming and outgoing photons along the normal and along the plane of the mirror, we see that there is no change of direction and hence momentum of the photons along the plane of the mirror. So the light beam does not exert force along the plane of the mirror. The component of the momentum of incoming photons along the normal is equal and opposite to the momentum of outgoing photons along the normal.
The component of the total momentum of the incoming photons per second along the normal is
=p'*cos 60° =p'/2
The component of the total momentum of the outgoing photons per second along the normal is
=-p'*cos 60° =-p'/2
Hence the change in momentum of the total photons falling per second along the normal
=p'/2 -(-p'/2) =p'
But the change in momentum per second is the force exerted. Hence the force exerted by the light beam on the mirror =p'
=1.0x10⁻⁸ N.
7. A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface.
ANSWER: The energy of photons falling per second on the plane surface =10 J.
The total momentum of these photons before they strike the plane surface,
p = E/c
=(10 J)/(3.0x10⁸ m/s)
=(10/3)x10⁻⁸ kg-m/s
70% of the light is absorbed by the surface, so after the strike, 70% of the momentum of photons becomes zero. The remaining 30% of the momentum has the same magnitude but opposite in direction because these photons are reflected back normally. So after the strike, the momentum of the photons falling per second is,
=0.70P*0 +(-0.30P)
=-0.30P
Hence the change in momentum per second,
=P -(-0.30P)
=1.30P
=1.30*(10/3)x10⁻⁸ kg-m/s
=4.3x10⁻⁸ kg-m/s
The force exerted by the incident beam on the surface is equal to the change in the momentum of the photons per second, i.e.,
=4.3x10⁻⁸ N.
8. A totally reflecting, small plane mirror placed horizontally faces a parallel beam of light as shown in figure (42-E1). The mass of the mirror is 20 g. Assume that there is no absorption in the lens and that 30% of the light emitted by the source goes through the lens. Find the power of the source needed to support the weight of the mirror. Take g =10 m/s².  |
| The figure for Q-8 |
ANSWER: To support the weight of the mirror, the parallel beam of light should apply a force equal to the weight of the mirror, i.e.
F =mg
=(20/1000)*10 N
=0.20 N.
Suppose the momentum of the photons falling per second on the mirror =p.
Since the mirror is totally reflecting, the momentum of the reflected photons =-p.
So the rate of change of momentum of the incident photons =p -(-p) =2p.
Since the rate of change of momentum = force applied,
F =2p
→p =F/2
=0.20/2 kg-m/s
=0.10 kg-m/s.
Let the power of the source =P watt, i.e. P joules per second. The energy of light passing through the lens per second,
E =0·30P J.
The relation between energy and momentum of photons is,
p =E/c
→E =pc
→0.30P =0.10*3.0x10⁸
→P =1.0x10⁸ W
=100 MW.
9. A 100 W light bulb is placed at the center of a spherical chamber of radius 20 cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. Find the pressure exerted by the light on the surface of the chamber.
ANSWER: Radius, r =20 cm =0.20 m.
Area of the sphere =4πr²
→A =4π*(0.20)² m²
=0.16π m²
Since the bulb is placed at the center of the sphere, the emitted photons will fall normally on the absorbing surface. Though the direction of photons will be different, it may be calculated as if falling on a normal surface.
The energy of photons emitted per second,
E =0.60*100 J/s =60 J/s.
The momentum of photons falling per second,
p =E/c
=60/3x10⁸ kg-m/s
=2x10⁻⁷ kg-m/s
Since the final momentum is zero due to absorption, the change in momentum per second,
= p =2x10⁻⁷ kg-m/s
Hence the magnitude of the total force on the spherical surface,
F =2x10⁻⁷ N
Hence the pressure exerted by the light on the spherical chamber,
=F/A
=2x10⁻⁷/0.16π N/m²
=4.0x10⁻⁷ Pa.
10. A sphere of radius 1.00 cm is placed in the path of a parallel beam of light of a large aperture. The intensity of the light is 0.50 W/cm². If the sphere completely absorbs the radiation falling on it, find the force exerted by the light beam on the sphere.
ANSWER: The force on the sphere will be the same as the force on a disc of radius 1.00 cm in the path of light. The area of this disc,
A =π*1² cm²
=π cm².
The intensity of light, I =0.50 W/cm²
So per cm², 0.50 J/s energy of photons are falling. The momentum of photons falling per second on a square cm,
=0.50/3x10⁸ kg-m/s
=1.67x10⁻⁹ kg-m/s.
The momentum of photons falling on the cross-section of the sphere per second,
p =1.67x10⁻⁹*π kg-m/s
=5.2x10⁻⁹ kg-m/s.
Since the sphere absorbs the light falling on it, the momentum of photons after absorption is zero. So the force exerted by the light beam on the sphere is the change of momentum of falling photons per second, which is
=p -0
=p
=5.2x10⁻⁹ N.
11. Consider the situation described in the previous problems. Show that the force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing.
ANSWER: Let us take the radius of the sphere = r and the intensity of the light = I. In the figure below, OX is the axis of the sphere parallel to the falling beam. P is a point on the sphere such that radius OP makes an angle θ from OX. If OP is moved through a very small angle dθ to point Q, then PQ =rdθ.  |
| Diagram for Q-11 |
Suppose PQ is moved on the surface of the sphere at the same inclination around OX, we get a ring with width rdθ. The area of this ring,
dA = rdθ*2πrSinθ =2πr²Sinθdθ.
The rays make an angle θ with radius OP, hence even after reflection, the rays will make an angle θ with the OP because the radius is perpendicular to the surface at P.
Consider a small area 'S' on this ring. Light energy falling on this area per second,
=S cosθ*I =SI cosθ.
The momentum of light falling on area S
=SI cosθ/c.
Since the light is reflected through the same angle from OP, the change of momentum of light energy per second is,
=2(SI cosθ/c)*cosθ
=2SI cos²θ/c, along PO.
Hence the force on this small area =2SIcos²θ/c.
Component of this force along the parallel beam (which is the force on the sphere for a small area S)
=(2SI cos²θ/c)*cosθ
=2SI cos³θ/c.
Hence the force on the ring of which part is S, {Substituting S with the area of the ring dA}
dF =(2dAIcos³θ/c)
=(4πr²I/c) cos³θSinθdθ.
For the total force on the sphere, we integrate it for the area of the exposed hemisphere, i.e. from θ =0 to θ =π/2.
F = ∫dF
=(4πr²I/c)∫cos³θ.sinθdθ
Let y =cosθ, dy =-sinθdθ. The limit of integration will be from, cos0 =1 to cosπ/2 =0. Hence
F =-(4πr²I/c)∫y³dy
=-(4πr²I/c)[y⁴/4]
=-(4πr²I/c)[0 -1/4]
=πr²I/c. ---------------- (i)
In the previous problem, if the intensity is I', the force on the sphere is
=(I'/c)*πr²
=πr²I'/c. ---------------- (ii)
(i) is derived for a fully reflecting sphere while (ii) is for a fully absorbing sphere. Our given sphere is partly absorbing. Suppose out of 0.50 W/cm² intensity (in the previous problem) I watt/cm² is fully reflected and the rest I' watt/cm² is fully absorbed. i.e. I +I' =0.50 W/cm², So the total force on the sphere
F =πr²I/c +πr²I'/c
=πr²(I+I')/c
=π*1²*0.50/3x10⁸ N
=5.2x10⁻⁹ N.
So the force on the sphere is the same even if the sphere is not fully absorbing.
12. Show that it is not possible for a photon to be completely absorbed by a free electron.
ANSWER: A free electron in a metal is attracted by the other molecules and nucleus of the atom. To bring it out of the metal a minimum amount of energy is needed which is called the work function φ of the metal. When a photon of energy hv is used to bring out the free electron from a metal surface, a part of photon energy hv is used as a work function. So the electron can only absorb the rest of the energy hv-φ and this energy shows up as the kinetic energy of the electron. Thus it is not possible for a photon to be completely absorbed by a free electron.
13. Two neutral particles are kept 1 m apart. Suppose by some mechanism some charge is transferred from one particle to the other and the electric potential energy lost is completely converted into a photon. Calculate the longest and the next smaller wavelength of the photon possible.
ANSWER: Since both the particles are neutral, if the -q charge is transferred from one particle to another, the first particle will have +q charge while the other will have -q charge. For a separation r, the electric potential energy between these two charges,
U =kq²/r.
For the same energy of a photon let its wavelength be equal to λ, then
U =hc/λ.
Equating the two energies,
kq²/r = hc/λ.
→λ =hcr/kq².
In this expression h, c, r, and k are constants. For the wavelength λ to be maximum, q must be minimum. And the minimum possible value of a charge is the charge of an electron.
So λₘₐₓ =hcr/ke²
=6.63x10⁻³⁴*3x10⁸*1/{9x10⁹*(1.6x10⁻¹⁹)²} m
=0.86x10³ m
=860 m.
The next greater charge will give the next smaller wavelength. And the next greater charge will be q =2e. Hence,
λ' =hcr/{k(2e)²}.
=6.63x10⁻³⁴*3x10⁸*1/{9x10⁹*4*2.56x10⁻³⁸} m
=0.215x10³ m
=215 m.
14. Find the maximum kinetic energy of the photoelectrons ejected when the light of wavelength 350 nm is incident on a cesium surface. The work function of Cesium =1.9 eV.
ANSWER: The energy of a photon of the light of wavelength λ =350 nm
E =hc/λ
=4.14x10⁻¹⁵x3x10⁸/350x10⁻⁹ eV
=3.5 eV
The work function of Cesium,
φ =1.9 eV
Hence the maximum kinetic energy of the emitted photoelectron,
= E-φ
=3.5 -1.9 eV
=1.6 eV.
15. The work function of a metal is 2.5x10⁻¹⁹ J. (a) Find the threshold frequency for photoelectric emission. (b) If the metal is exposed to a light beam of frequency 6.0x10¹⁴ Hz. What will be the stopping potential?
ANSWER: The work function,
φ =2.5x10⁻¹⁹ J.
(a) Threshold frequency 𝜈 of photons is the minimum frequency for which photoelectrons are just able to come out of the metal. The energy of this photon is equal to the work function. So,
h𝜈 =φ
→𝜈 =φ/h
=2.5x10⁻¹⁹/6.63x10⁻³⁴ Hz
=3.8x10⁻¹⁴ Hz.
(b) Frequency of light beam,
𝜈 =6.0x10¹⁴ Hz
If the stopping potential is Vₒ, then
eVₒ =Kₘₐₓ =h𝜈 -φ
→Vₒ =h𝜈/e -φ/e
=6.63x10⁻³⁴*6.0x10¹⁴/1.6x10⁻¹⁹ -2.5x10⁻¹⁹/1.6x10⁻¹⁹ volts
=2.48 -1.56 volts
=0.92 volts.
16. The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.
ANSWER: (a) Given φ =4.0 eV,
→φ =4.0 eV.
Let the threshold wavelength =λ,
Then hc/λ =φ
→λ =hc/φ
=4.14x10⁻¹⁵*3x10⁸/4.0 m
=3.10x10⁻⁷ m
=310x10⁻⁹ m
=310 nm.
(b) Vₒ =2.5 V,
Let the corresponding wavelength of light for this stopping potential be λ. Then,
eVₒ =hc/λ -φ
→2.5=4.14x10⁻¹⁵*3x10⁸/λ-4.0 eV
→6.5 =12.42x10⁻⁷/λ
→λ =1.90x10⁻⁷ m
=190x10⁻⁹ m
=190 nm.
17. Find the maximum magnitude of the linear momentum of a photoelectron emitted when the light of wavelength 400 nm falls on a metal having work function 2.5 eV.
ANSWER: λ =400 nm
→λ =4x10⁻⁷ m
φ =2.5 eV =2.5x1.6x10⁻¹⁹ J
The maximum kinetic energy of the photoelectron,
E =hc/λ -φ
=6.63x10⁻³⁴*3x10⁸/4x10⁻⁷ -2.5x1.6x10⁻¹⁹
=5.0x10⁻¹⁹ -4.0x10⁻¹⁹ J
=1.0x10⁻¹⁹ J
If p is the linear momentum of the photoelectron then,
E = p²/2m
→p² =2mE
=2x9.1x10⁻³¹*1.0x10⁻¹⁹
=18.2x10⁻⁵⁰
→p =4.2x10⁻²⁵ kg-m/s.
18. When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photocurrent. Find the threshold wavelength for the metal.
ANSWER: The stopping potential,
Vₒ =1.1 volts
Wavelength, λ =400 nm =4x10⁻⁷ m
If φ is the work function then.
eVₒ =hc/λ -φ
→eVₒ =hc/λ -hc/λ'
{Where λ' is the threshold wavelength.}
→hc/λ' =hc/λ -eVₒ
→1/λ' =1/λ -eVₒ/hc
=1/4x10⁻⁷ -1.6x10⁻¹⁹*1.1/{6.63x10⁻³⁴*3x10⁸}
=2.5x10⁶ -0.88x10⁶
→1/λ' =1.62x10⁶
→λ' =6.20x10⁻⁷ m
=620x10⁻⁹ m
=620 nm.
19. In an experiment on the photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. The data collected are as follows:
wavelength (nm): 350 400 450 500 550
stopping potential (V):1.45 1.00 0.66 0.38 0.16
Plot the stopping potential against the inverse of wavelength (1/λ) on graph paper and find (a) the Plank constant, (b) the work function of the emitter, and (c) the threshold wavelength.
ANSWER: The relation between stopping potential and the wavelength of the incident light is,
eVₒ +φ =hc/λ
→1/λ =eVₒ/hc +φ/hc
This is in the form of y =mx+C, which is an equation of a straight line.
Here, y = 1/λ, x =Vₒ,
m =e/hc, {Slope of the line},
C =φ/hc, {Intercption of the line on the Y-axis}.
Let us plot the graph. We have,
1/λ (*10⁶)=2.86 2.50 2.22 2.00 1.82
Vₒ (V) = 1.45 1.00 0.66 0.38 0.16
 |
| Graph for Q-19 |
From the graph, C =φ/hc =1.72x10⁶ m⁻¹
→φ =h*3x10⁸*1.72x10⁶ J
=5.16x10¹⁴*h J -------------- (i)
(a) Slope of the graph m =tanθ =1.14x10⁶/1.45
→e/hc =0.8x10⁶
→h =ex10⁻⁶/(0.8c) J-s
=1.0x10⁻⁶/(0.8c) eV-s
=1.0x10⁻⁶/(0.8x3x10⁸) eV-s
=4.2x10⁻¹⁵ eV-s.
(b) From (i), work function
φ = 5.16x10¹⁴ h
=5.16x10¹⁴*4.2x10⁻¹⁵ eV
=2.16 eV.
(c) Threshold wavelength λ is given as
hc/λ =φ
→λ =hc/φ
=4.2x10⁻¹⁵*3x10⁸/2.16 m
=5.83x10⁻⁷ m
=583x10⁻⁹ m
=583 nm.
{Note: Accuracy of the result depends upon accuracy of plotting and eye estimation}
20. The electric field associated with a monochromatic beam becomes zero 1.2x10¹⁵ times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.
ANSWER: The electric field of an electromagnetic wave becomes zero two times in a cycle. So in a second, it becomes zero twice its frequency. Thus the frequency of the given monochromatic beam is half of the given number.
Frequency 𝜈 =6.0x10¹⁴
The maximum kinetic energy of the photoelectrons, when this beam falls on a metal surface of work function φ =2.0 eV, is
=h𝜈 -φ
=4.14x10⁻¹⁵*6x10¹⁴-2.0 eV
=2.48 -2.0 eV
=0.48 eV.
21. The electric field associated with a light wave is given by
E =Eₒsin[(1.57x10⁷ m⁻¹)(x -ct)].
Find the stopping potential when the light is used in an experiment on the photoelectric effect with the emitter having work function 1.9 eV.
ANSWER: The equation for the electric field in a wave is written as,
E =EₒSin ⍵(x/c -t)
=Eₒ Sin (⍵/c)(x -ct)
Comparing to the given equation, we get,
⍵/c =1.57x10⁷ rad/s
So the frequency 𝜈 is given as,
2π𝜈/c = 1.57x10⁷ rad/s
→𝜈 =1.57x10⁷*3x10⁸/2π Hz
=7.5x10¹⁴ Hz
The stopping potential Vₒ is given as
eVₒ =h𝜈 -φ
→Vₒ =(6.63x10⁻³⁴*7.5x10¹⁴-1.9*1.60x10⁻¹⁹)/1.6x10⁻¹⁹ volt
=3·1 -1·9 volt
=1·2 V.
22. The electric field at a point associated with a light wave is
E =(100 V/m) sin[(3x10¹⁵ s⁻¹)t] sin[(6x10¹⁵ s⁻¹)t].
If this light falls on a metal surface having a work function of 2.0 eV, what will be the maximum kinetic energy of the photoelectrons?
ANSWER: From,
Cos(A-B)-Cos(A+B) =2SinA.SinB,
The given equation is,
E =(100 V/m)*½[(Cos 3x10¹⁵ s⁻¹t)-(Cos 9x10¹⁵ s⁻¹t)]
So there are two waves mixed in the incident light having angular frequencies ⍵ =3x10¹⁵ s⁻¹ and 9x10¹⁵ s⁻¹.
The maximum kinetic energy of a photoelectron will be related to the maximum frequency of the incident light. So we take ⍵ =9x10¹⁵ s⁻¹.
→2π𝜈 =9x10¹⁵ s⁻¹
→𝜈 =1.43x10¹⁵ Hz.
So the maximum kinetic energy of a photoelectron,
=h𝜈 -φ
If we take the unit as eV,
φ =2.0 eV, h =4.14x10⁻¹⁵ eV-s
So, Kₘₐₓ =4.14x10⁻¹⁵*1.43x10¹⁵-2.0 eV
=5·92 -2·0 eV =3·92 eV.
23. A monochromatic light source of intensity 5 mW emits 8x10¹⁵ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 eV Calculate the work function of the metal.
ANSWER: Energy emitted by light per second is the intensity of light. So in the given problem, 8x10¹⁵ photons have energy =5 mW =5x10⁻³ J.
So the energy of each photon,
h𝜈 =5x10⁻³/8x10¹⁵ J
=6.25x10⁻¹⁹ J
=3.9 eV
Stopping potential in terms of eV,
Vₒ =h𝜈 -φ
→2.0 =3.9 -φ
→φ =3.9 -2.0 eV
=1·9 eV.
24. Figure (42-E2) is the plot of the stopping potential versus the frequency of the light used in an experiment on the photoelectric effect. Find (a) the ratio h/e and (b) the work function.  |
| Figure for Q-24 |
ANSWER: (a) The stopping potential is given as,
eV。=hv -φ
→V。=hv/e -φ/e ------ (i)
It is an equation of a straight line in the form of
y =mx +C
The slope of the line,
m =tan θ =h/e,, and the intercept on the y-axis, C =φ/e.
From the graph,
m =1.656/4x10¹⁴ V-s
→h/e =4·14x10⁻¹⁵ V-s.
(b) From equation (i),
V。=(h/e)v -φ/e
Taking V。=1.656 V and v =5x10¹⁴ Hz
from the graph,
1.656 =4.14x10⁻¹⁵*5x10¹⁴ -φ/e
→φ/e =2.07 -1.656
→φ =0·414*e J
=0·414 eV.
25. A photographic film is coated with a silver bromide layer. When light falls on this film silver bromide molecules dissociate and the film records the light there. A minimum of 0.6 eV is needed to dissociate a silver bromide molecule. Find the maximum wavelength of light that can be recorded by the film.
ANSWER: From the problem, we conclude that the minimum energy of a photon needed to dissociate a silver bromide is 0.6 eV. It will correspond to the maximum wavelength 𝜆 of the light. So,
hc/𝜆 =0.6 eV
→6.63x10⁻³⁴*3x10⁸/𝜆 =0.6*1.6x10⁻¹⁹ J
→1.99x10⁻²⁵/𝜆 =9.6x10⁻²⁰
→𝜆 =2.070x10⁻⁶ m =2070 nm.
26. In an experiment on the photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches its saturation value. Assuming that on average one out of every 10⁶ photons is able to eject a photoelectron, find the photocurrent in the circuit.
ANSWER: The energy of each photon of the incident light,
E =hc/𝜆
=6.63x10⁻³⁴*3x10⁸/(400x10⁻⁹) J
≈5x10⁻¹⁹ J.
The total energy of the light =5.0 J/s.
Number of photons moving in the light beam,
=5.0/5x10⁻¹⁹ per/s
=1x10¹⁹ per/s.
Since one out of 10⁶ photon is able to eject a photoelectron, the number of photoelectrons emitting out per second,
N =1x10¹⁹/10⁶
=1x10¹³.
Hence the current in the circuit,
i =N*e A
=1x10¹³*1.6x10⁻¹⁹ A
=1.6x10⁻⁶ A
=1.6 µA.
27. A silver ball of radius 4.8 cm is suspended by a thread in a vacuum chamber. Ultraviolet light of wavelength 200 nm is incident on the ball for some time during which a total light energy of 1.0x10⁻⁷ J falls on the surface. Assuming that on average one photon out of every ten thousand is able to eject a photoelectron, find the electric potential at the surface of the ball assuming zero potential at infinity. What is the potential at the center of the ball?
ANSWER: The wavelength of the incident ultraviolet light,
λ =200 nm =2x10⁻⁷ m.
The energy of each photon,
E =hc/λ
=6.63x10⁻³⁴*3x10⁸/2x10⁻⁷ J
≈1.0x10⁻¹⁸ J.
Total number of photons that fell on the ball,
N =1.0x10⁻⁷ J/1.0x10⁻¹⁸ J
=1.0x10¹¹.
The number of ejected photoelectrons,
N'=1.0x10¹¹/10⁴
=1.0x10⁷.
The positive charge on the ball due to ejected photoelectrons,
q =1.0x10⁷*1.6x10⁻¹⁹ C
=1.6x10⁻¹² C.
The electric potential on the surface of the ball,
V =Kq/r
K =9x10⁹ N-m²/C²,
r =4.8 cm =0.048 m
Hence V =9x10⁹*1.6x10⁻¹²/0.048 volt
=0·30 volts.
Since the electric potential inside a sphere remains constant and equal to the surface potential, the electric potential at the center of the ball =0·30 volts.
28. In an experiment on the photoelectric effect, the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell (figure 42-E3). A magnetic field B exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.  |
| Figure for Q-28 |
ANSWER: The value of B will be minimum when the emitted electrons having the maximum kinetic energy fail to reach the collector plate. Electrons having maximum kinetic energy will be emitted by the light having minimum wavelength i.e. λ =400 nm.
The maximum kinetic energy of the photoelectron,
K =hc/λ -φ
=4.14x10⁻¹⁵*3x10⁸/400x10⁻⁹ -2.39 eV
=3.105 -2.39 eV
=0.715 eV.
Since the emitted electrons encounter a magnetic field B that is perpendicular to its direction of motion, they experience a force perpendicular to both B and v and it results in a circular motion of the photoelectrons. If the radius of this circular path is just equal to the separation of plates, there will be no current in the circuit. The radius is given as,
r =mv/qB
Here, r =d and mv =√(2mK)
So, d =√(2mK)/qB
→B =√(2mK)/qd
=√(2*9.1x10⁻³¹*0.715*1.6x10⁻¹⁹)/(1.6x10⁻¹⁹*0.10)
=2.85x10⁻⁵ T.
29. In the arrangement shown in figure (42-E4). y =1.0 mm, d =0.24 mm and D =1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.  |
| Figure for Q-29 |
ANSWER: The fringe width,
w=2y=2*1.0 mm=2.0 mm=2.0x10⁻³ m.
D =1.2 m, d =0.24 mm =2.4x10⁻⁴ m.
The wavelength of light, 𝜆 =wd/D.
→𝜆 =2x10⁻³*2.4x10⁻⁴/1.2 m
=4x10⁻⁷ m.
The energy of a photon, E =hc/𝜆
→E =4.14x10⁻¹⁵*3x10⁸/4x10⁻⁷ eV
=3.1 eV.
The work function of the emitter, φ =2.2 eV.
Stopping potential V' needed to stop the photocurrent is given as,
eV' =E -φ =3.1 -2.2 =0.9 eV
→V' ={0.9*e/e} V = 0·9 V.
30. In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (φ =4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector.
ANSWER: When photons are incident on the emitter surface, the electrons in the metal absorbing the photons are detached from the molecules but they may either lose some or all their kinetic energy with collisions with other molecules in the metal. So the electrons coming out of the emitter surface may have kinetic energies varying from zero to a maximum K =hc/λ-φ.
→K =4.14x10⁻¹⁵*3x10⁸/(200x10⁻⁹) -4.5 eV
→K =6.21 -4.5 eV =1.71 eV.
Since here the collector plate is at 2.0 V with respect to the emitter plate, the kinetic energy of 2.0 eV will be added to the electrons coming out of the emitter plate. The electrons coming out of the emitter plate have already kinetic energy from zero to 1.71 eV. Thus the electrons reaching the collector plate will have minimum kinetic energy =0+2.0 eV =2.0 eV and the maximum kinetic energy =1.71+2.0 eV =3.71 eV.
31. A small piece of cesium metal (φ =1.9 eV) is kept at a distance of 20 cm from a large metal plate having a charge density of 1.0x10⁻⁹ C/m² on the surface facing the cesium piece. A monochromatic light of wavelength 400 nm is incident on the cesium piece. Find the minimum and maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in the electric field due to the small piece of cesium present.
ANSWER: Given λ =400 nm and φ =1.9 eV. Hence without any external force, the emitted photoelectrons will have kinetic energy varying from zero to K =hc/λ-φ,
→K=4.14x10⁻¹⁵*3x10⁸/400x10⁻⁹-1.9 eV
=3.1 -1.9 eV =1.2 eV.
Here we have an accelerating force for the emitted photoelectrons due to the large charged plate. The electric field near a charged plate is E =𝜎/εₒ, where 𝜎 is the charge density. Given,
𝜎 =1.0x10⁻⁹ C/m² and we know,
εₒ =8.85x10⁻¹² C²/N-m².
So, E =1.0x10⁻⁹/8.85x10⁻¹² N/C
=113 N/C.
The electric potential at the cesium metal due to the charged plate,
V =E*d
=113*0.20 V
=22.6 V.
This will impart a kinetic energy of 22.6 eV to the emitted photoelectrons near the cesium plate. So the minimum kinetic energy of the photoelectrons reaching the large metal plate is =0+22.6 eV =22·6 eV and the maximum kinetic energy =1.2 +22.6 eV =23·8 eV.
32. Consider the situation of the previous problem. Consider the fastest electron emitted parallel to the large metal plate. Find the displacement of this electron parallel to its initial velocity before it strikes the large metal plate.
ANSWER: The fastest electron emitted parallel to the large metal plate will have velocity v parallel to the plate and its kinetic energy K =1.2 eV, (as calculated in the previous problem). But its velocity perpendicular to the plates is zero.
So v =√(2K/m)
→v =√(2*1.2*1.6x10⁻¹⁹/9.1x⁻³¹) m/s
=6.5x10⁵ m/s.
Now it is a problem like a stone thrown parallel to the ground.  |
| Diagram for Q-32 |
Force on the electron perpendicular to the plate due to the electric field, E =113 N/C,
F =eE, and the acceleration along this force,
a =F/m =eE/m
=1.6x10⁻¹⁹*113/9.1x10⁻³¹ m/s
=2x10¹³ m/s².
If t is time to reach the large plate, then
d =½at²
→t =√(2d/a) s
=√(2*0.20/2x10¹³) s
=1.41x10⁻⁷ s
Hence the displacement of this electron parallel to its initial velocity before it strikes the large metal plate,
S =vt
=6.5x10⁵*1.41x10⁻⁷ m
=0.092 m
=9·2 cm.
33. A horizontal cesium plate (φ =1.9 eV) is moved vertically downward at a constant speed v in a room full of radiation of wavelength 250 nm and above. What should be the minimum value of v so that the vertically upward component of velocity is nonpositive for each photoelectron?
ANSWER: When the plate is unmoved, the maximum upward velocity v' of an emitted photoelectron will correspond to the maximum possible kinetic energy K of the emitted photoelectron. From Einstien's photoelectric equation,
K =hc/λ -φ
=4.14x10⁻¹⁵*3x10⁸/250x10⁻⁹ -1.9 eV
=5.0 -1.9 eV
=3.1 eV.
Velocity v' =√(2K/m)
→v'=√(2*3.1*1.6x10⁻¹⁹/9.1x10⁻³¹) m/s
=1.04x10⁶ m/s.
When the plate is moved vertically downward with a velocity v, each emitted photoelectron gets additional velocity v. Net velocity of the fastest vertically upward moving electron,
=v' -v.
For the vertically upward component of the velocity of emitted photoelectrons to be nonpositive,
v' -v =0
→v =v' =1·04x10⁶ m/s.
34. A small metal plate (work function φ) is kept at a distance d from a singly ionized, fixed ion. A monochromatic light beam is incident on the metal plate and photoelectrons are emitted. Find the maximum wavelength of the light beam so that some of the photoelectrons may go around the ion along a circle.
ANSWER: The photoelectrons going around the ion must have a force of attraction as centripetal force. Thus the charge on the ion is positive e so that the electrostatic force between the ion and the electron provides this force. The centripetal electrostatic force,
F =Ke²/d²
When the photoelectrons having the maximum kinetic energy have a velocity parallel to the emitter plate and have centrifugal force just equal to F, they will go around the ion. The maximum kinetic energy of a photoelectron,
E =hc/λ -φ
Its velocity, v =√(2E/m)
The centrifugal force =mv²/d
=m*2E/md
=2E/d
=2(hc/λ -φ)/d
Equating the two forces,
Ke²/d² =2(hc/λ -φ)/d
Ke²/d =2(hc/λ -φ)
→hc/λ =Ke²/2d +φ =(Ke²+2φd)/2d
→λ =2dhc/(Ke²+2φd)
=2dhc/{e²/4πεₒ +2φd}
=2dhc/{(e² +8πεₒφd)/4πεₒ}
=8πεₒdhc/(e² +8πεₒφd).
35. A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. (a) A particular electron absorbs a photon and makes two collisions before coming out of the metal. Assuming that 10% of the extra energy is lost to the metal in each collision, find the kinetic energy of this electron when it comes out of the metal. (b) Under the same assumptions, find the maximum number of collisions the electrons can suffer before they become unable to come out of the metal.
ANSWER: (a) λ =400 nm, φ =2.2 eV.
The energy of the photon is,
E = hc/λ
=4.14x10⁻¹⁵*3x10⁸/400x10⁻⁹ eV
=3.1 eV.
This energy of the photon is absorbed by the electron and gains kinetic energy. The energy lost by the absorbing electron in one collision is 10%, hence remaining energy after two collisions,
=(0.9)²*3.1 eV =2.51 eV
The electron will lose energy equal to the work function in coming out of the metal. So the kinetic energy of this electron after coming out =2.51 -φ
=2.51 -2.2 eV =0.31 eV.
(b) Remaining kinetic energy after two collisions =2.51 eV.
After the third collision remaining energy =0.9*2.51 eV =2.26 eV.
After the fourth collision remaining energy =0.9*2.26 eV =2.03 eV.
The remaining energy is now less than the work function and it will not be able to come out of the metal.
The electron will have 4 number of collisions before it is unable to come out of the metal.
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