Saturday, April 15, 2023

H C Verma solutions, BOHR'S MODEL AND PHYSICS OF THE ATOM, Chapter-43, EXERCISES, Q31 TO Q46, Concepts of Physics, Part-II

Bohr's Model and Physics of the Atom


Exercises, Q31 to Q46


     31.  A beam of monochromatic light of wavelength 𝜆 ejects photoelectrons from a cesium surface (Ჶ =1.9 eV). These photoelectrons are made to collide with hydrogen atoms in the ground state. Find the maximum value of 𝜆 for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.  


ANSWER: (a) To ionize a hydrogen atom in the ground state, 13.6 eV of energy is needed. The emitted electrons from the cesium surface must possess 13.6 eV of kinetic energy. Due to the work function of the cesium surface, the minimum energy of the photons of the monochromatic light must be equal to 13.6 eV +1.9 eV =15.5 eV and in this case, the 𝜆 will have the maximum value.

   Thus, for the maximum wavelength 𝜆, the energy of photons of monochromatic light should be

hc/𝜆 =15.5 eV

→𝜆 =hc/15.5

    =4.14x10⁻¹⁵*3x10⁸/15.5 m

    =8.0x10⁻⁸ m

    =80x10⁻⁹ m

    =80 nm.


(b) To get excited from the ground state to the first excitation state, a hydrogen atom needs energy equal to

E =13.6(1 -1/2²)

   =10.2 eV

So the emitted electrons from the cesium surface need a minimum kinetic energy of 10.2 eV. In this case, the maximum value of 𝜆 i.e. minimum energy of the photons should be equal to 10.2 eV +1.9 eV =12.1 eV.

So, hc/𝜆 =12.1

→𝜆 =hc/12.1

     =4.14x10⁻¹⁵*3x10⁸/12.1 m

     =1.02x10⁻⁷ m

      =102x10⁻⁹ m 

      =102 nm.   


(c) The excited hydrogen atom emits visible light when electrons jump down to the n =2 state from higher states. So, the minimum kinetic energy of the electrons emitted from the cesium surface should be such that it takes the electron in the ground state of the hydrogen atom to n =3 state. This minimum energy =13.6{1 -1/3²) eV 

   =12.1 eV

So the energy of the photons of the monochromatic light should be at least

=12.1+1.9 eV

=14.0 eV

Hence

hc/𝜆 =14.0

→𝜆 =4.14x10⁻¹⁵*3x10⁸/14.0 m

     =8.9x10⁻⁸ m

     =89x10⁻⁹ m

     =89 nm.

  





 

     32.  Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0 m. The electrons are now scattered by an atomic hydrogen sample in the ground state. What should be the minimum value of E so that red light of wavelength 656.3 nm may be emitted by the hydrogen?  


ANSWER: The wavelength of the given red light

𝜆 =656.3 nm. 

The energy of a photon of this wavelength,

E' =hc/𝜆

  =4.14x10⁻¹⁵*3x10⁸/656.3x10⁻⁹ eV

  =1.9 eV.

The red light is in the visible range and visible range lights are emitted in a hydrogen atom when the electrons from higher states come down to the n =2 states. Since an electron in the ground state requires energy =13.6*(1-1/2²) =10.2 eV to reach the n =2 state, the accelerated electrons must possess 10.2 +1.9 =12.1 eV energy to make the electrons in the ground state to jump to the required excited state.

 Since the electrons are accelerated through 1.0 m in an electric field E, the work done on it (=eE joules =E eV) will be the kinetic energy of the electrons. Equating it with the energy needed we get, 

 E =12.1 V/m

   

   




 

     33.  A neutron having a kinetic energy of 12.5 eV collides with a hydrogen atom at rest. Neglect the difference in mass between the neutron and the hydrogen atom and assume that the neutron does not leave its line of motion. Find the possible kinetic energies of the neutron after the event.  


ANSWER: We know that when two bodies of equal masses elastically collide in a linear direction, their velocities interchange. So the neutron in this case will have zero velocity after the collision and hence zero kinetic energy.

  This can also be proved mathematically. The neutron does not leave its line of motion after the collision which means that the hydrogen atom at rest will also move along the same line after the collision (from the conservation of momentum in a direction principle).  

    Before collision let the momentum of the neutron =p and that of the hydrogen atom =0. After collision let the momenta of neutron = p' and that of hydrogen atom =p". So from the momentum conservation principle, 

p+0 =p'+p" 

→p' +p" =p  ------------------ (i)

The kinetic energies of the neutron and hydrogen atom are K (=12.5 eV) and zero before the collision. 

 

If the kinetic energies of neutron and hydrogen atoms are K' and K" respectively after the collision, then from the energy conservation rule

K'+K" =K +0. 

Putting the expression of kinetic energy in terms of momentum

p'²/2M +p"²/2M =p²/2M

→p'² +p"² =p²  ------ (ii)

But from (i)

(p' +p")² =p²

→p'² +p"² +2p'p" =p²

→p²+2p'p" =p²     {from (ii)}

→2p'p" =0 

→p'p" =0 

After the collision, the hydrogen atom will have some velocity and momentum, p" ≠0 which requires p' to be zero. 

  So the neutron will have zero momentum after the collision. 






 

     34.  A hydrogen atom moving at speed v collides with another hydrogen atom kept at rest. Find the minimum value of v for which one of the atoms may get ionized. The mass of a hydrogen atom =1.67x10⁻²⁷ kg.   


ANSWER: Let after the collision, the velocities of hydrogen atoms are v' and v". From the momentum conservation principle, 

Mv =Mv' +Mv" 

→v =v' +v"   -------- (i). 

If the energy required for the ionization of one hydrogen atom is E, then from the conservation of energy principle, 

½Mv² =½Mv'²+½Mv"²+E 

 →v² =v'²+v"²+2E/M 

But by squaring both sides of (i), we get 

v² =v'²+v"²+2v'v"  

By comparison, 

2E/M =2v'v"

→E/M =v'v" 

Now, 

(v'-v")² =(v'+v")²-4v'v" 

→(v'-v")² = v² -4E/M 

→v² =(v'-v")²+4E/M 

For v to be minimum, 

(v'-v")² =0 

→v' =v"

That is both atoms move at the same speed after collision. Also, for the minimum v, 

v² =4E/M

We know, E =13.6 eV. Hence, 

v²=4*13.6x1.6x10⁻¹⁹/1.67x10⁻²⁷ 

   =5.24x10⁹

→v =7.2x10⁴ m/s.             





  

 

     35.  A neutron moving with a speed v strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen =1.67x10⁻²⁷ kg. 


ANSWER: From the conservation of momentum principle, both will have the same speed v after the collision but opposite in direction. It can be shown below. Let after the collision the speed of the neutron is v' and that of the hydrogen atom be v". Then from the momentum conservation principle, 

Mv -Mv =Mv'+Mv"   

→v' =-v" 

So both have equal and opposite speeds.   

   Suppose an energy E is absorbed by the hydrogen atom in the collision, then from the conservation of energy principle, 

  ½Mv² *2=½Mv'²*2 +E 

→v² =v'² +E/M 

For v to be minimum, v' =0 which means that after the collision both come to rest and, 

v² =E/M 

Now the minimum amount of energy E that a hydrogen atom in the ground state can absorb is the energy required for the transition from n =1 to n =2 state i.e. equal to 10.2 eV. 

So, v² =10.2*1.6x10⁻¹⁹/1.67x10⁻²⁷ 

      =9.77x10⁸ m²/s²

  →v =3.13x10⁴ m/s.    





 

     36.  When a photon is emitted by a hydrogen atom, the photon carries momentum with it. (a) Calculate the momentum carried by the photon when a hydrogen atom emits light of wavelength 656.3 nm. (b) With what speed does the atom recoil during this transition? Take the mass of the hydrogen atom =1.67x10⁻²⁷ kg. (c) Find the kinetic energy of the recoil of the atom.  


ANSWER: (a) We have the wavelength 𝜆 =656.3 nm. 

The momentum of this photon,

p =h/𝜆, where h is the plank's constant.

 =6.63x10⁻³⁴/656.3x10⁻⁹ kg-m/s

 =1.0x10⁻²⁷ kg-m/s.


(b) From the conservation of momentum principle, the same amount of momentum will be gained by the atom but in the opposite direction. Let the final speed of the atom =v. Then,

Mv =p

→v =p/M

      =1.0x10⁻²⁷/1.67x10⁻²⁷ m/s

      =0.60 m/s.

 

(c) The kinetic energy of the recoiled atom,

  =½Mv²

  =½*1.67x10⁻²⁷*(0.6)² J

  =3.0x10⁻²⁸ J

  =1.9x10⁻⁹ eV.

  





 

     37.  When a photon is emitted from an atom, the atom recoils. The kinetic energy of the recoil and the energy of the photon comes from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from n =3 to n =2. Calculate the fractional change in the wavelength of the light emitted, due to the recoil. 


ANSWER: When we do not consider the recoil of the atom, the wavelength is given as, 

 1/𝜆 =1.097x10⁷{1/2² -1/3²}

→1/𝜆 =1.52x10⁶

→𝜆 =6.563x10⁻⁷ m

     =656.3x10⁻⁹ m

     =656.3 nm.

When we consider the recoil of the atom, the momentum gained in recoil by the atom is equal to the momentum of the photon in magnitude. This momentum,

p =h/𝜆

 =6.63x10⁻³⁴/656.3x10⁻⁹ kg-m/s

 =1.0x10⁻²⁷ kg-m/s

Hence the kinetic energy gained by the atom in recoil is

E =½Mv² =½(Mv)²/M

   =p²/2M

 =(1.0x10⁻²⁷)²/(2*1.67x10⁻²⁷) J

 =3.0x10⁻²⁸ J

 =1.9x10⁻⁹ eV.

In this case, the photon will have less energy by this amount of E. This revised energy of the photon will be

E' =13.6{1/2² -1/3²} -E

  =1.9 eV -1.9x10⁻⁹ eV

The wavelength of this light,

 𝜆' =hc/E' 

Hence the change in the wavelength   

𝜆' -𝜆 ={hc/(1.9-1.9x10⁻⁹)}-hc/1.9

=hc[(1.9-1.9+1.9x10⁻⁹)/{1.9*(1.9-1.9x10⁻⁹)}]

=hc[1.9x10⁻⁹/(1.9*(1.9-1.9x10⁻⁹)}

=hc*1.0x10⁻⁹/{1.9(1-10⁻⁹)}

=hc*1.0x10⁻⁹/1.9

Because 1-10⁻⁹ is almost equal to 1. 

Hence the fractional change in the wavelength

𝜆'-𝜆/𝜆 ={hc*1.0x10⁻⁹/1.9}/(hc/1.9)

  =1.9x10⁻⁹/1.9 

 =10⁻⁹.

 




 

     38.  The light emitted in the transition n =3 to n =2 in hydrogen is called Hₐ light. Find the maximum work function a metal can have so that Hₐ light can emit photoelectrons from it.  


ANSWER: For Hₐ light to emit photoelectrons from a metal surface, its photons should have energy equal to or more than the work function. Hence the maximum work function = energy of the photon of Hₐ light,

=13.6*{1/2² -1/3²} eV

=1.9 eV.

 




 

     39.  Light from the Balmer series of hydrogen is able to eject photoelectrons from a metal. What can be the maximum work function of the metal?  


ANSWER: Light from the Balmer series of hydrogen is emitted when electrons from higher states jump down to the n =2 state. The maximum work function of the metal from which this light is able to eject photoelectrons will be equal to the maximum energy of the photons. The maximum energy of a photon will be when the electron in the hydrogen atom jumps from n =∞ state down to n = 2 states. So the maximum work function,

φ =13.6{1/2² -1/∞²} eV

   =13.6/4 eV

   =3.4 eV.

 




 

     40.  Radiation from a hydrogen discharge tube falls on a cesium plate. Find the maximum possible kinetic energy of the photoelectrons. The work function of cesium is 1.9 eV.  


ANSWER: The maximum energy of a photon of radiation from the hydrogen discharge tube will be when the emission is from the transition of the electron from n =∞ to n =1 state in the hydrogen atom. The energy of this photon will be,

E =13.6{1/1² -1/∞²} eV

   =13.6 eV.

The work function of cesium φ =1.9 eV. Hence out of 13.6 eV of energy of the photon, 1.9 eV will be used to free the photoelectron from the metal surface and the rest will appear as the kinetic energy of the photoelectron. Hence the kinetic energy of the photoelectron. Hence the maximum possible kinetic energy of a photoelectron in this case is

  =13.6 eV -1.9 eV

  =11.7 eV.

 




 

     41.  A filter transmits only radiation of wavelength greater than 440 nm. Radiation from a hydrogen discharge tube goes through such a filter and is incident on a metal of work function 2.0 eV. Find the stopping potential that can stop the photoelectrons.   


ANSWER: The wavelength around 440 nm falls in the Balmer series. The photons just more than 440 nm will have the maximum energy and when they eject photoelectrons their stopping potential will be maximum. Let 𝜆 be the wavelength of such radiation. In the hydrogen atom electrons from higher states jump down to the n =2 state. Let us put the wavelength =440 nm as a trial.

1/(440 nm) =1.097x10⁷{1/2² -1/n²}

→1/4 -1/n² =1/{(440x10⁻⁹*1.097x10⁷) →1/n² =1/4 -1/4.83 =0.043

→n² =23.3

→n =4.82

But n should be an integer and for the wavelength to be just more than 440 nm it should be less than 4.82. So the electron jumps down from n =4 state to n =2 states.

The energy of this photon

E =13.6{1/2² -1/4²}

   =2.55 eV.

Given work function φ =2.0 eV

Let the stopping potential =Vₒ

Then from the Einstein equation,

E -φ =Vₒ

→Vₒ =2.55 eV -2.0 eV

       =0.55 eV.

 




 

     42.  The earth revolves around the sun due to gravitational attraction. Suppose that the sun and the earth are point particles with their existing masses and that Bohr's quantization rule for angular momentum is valid in the case of gravitation. (a) Calculate the minimum radius the earth can have for its orbit. (b) What is the value of the principal quantum number n for the present radius? The mass of the earth = 6.0x10²⁴ kg, the mass of the sun =2.0x10³⁰ kg, earth-sun distance =1.5x10¹¹ m. 


ANSWER: From Bohr's quantization rule, the possible orbits are those for which the angular momentum is an integral multiple of h/2π. 

Let v be the speed of the earth in the nth orbit. 

Then GMm/r² =mv²/r 

→r =GM/v² ------------------- (i) 

But from Bohr's quantization rule, 

mvr =nh/2π 

→v =nh/2πmr 

 Putting this expression in (i) 

r =GM(2πmr)²/n²h² 

→r =4π²GMm²r²/n²h² 

→r =n²h²/(4π²GMm²)


(a) The minimum radius that can be from this rule will be obtained by putting n =1. 

r =h²/(4π²GMm²) 

=(6.63x10⁻³⁴)²/(4π²6.67x10⁻¹¹*2x10³⁰*(6x10²⁴)²} m

=2.3x10⁻¹³⁸ m.


(b) Present radius R =1.5x10¹¹ m

Clearly, R =rn²

→n² =R/r

     =1.5x10¹¹/2.3x10⁻¹³⁸

     =6.5x10¹⁴⁸

→n =2.5x10⁷⁴. 


       




 

     43.  Consider a neutron and an electron bound to each other due to gravitational force. Assuming Bohr's quantization rule for angular momentum to be valid in this case, derive an expression for the energy of the neutron-electron system.  


ANSWER: Let the mass of the neutron =M and the mass of the electron =m. If the radius of the orbit =r then, 

GMm/r² =mv²/r, where v is the speed of the electron. 

→v² =GM/r. ----------------- (i) 

From Bohr's quantization principle, 

mvr =nh/2π  

→v =nh/2πmr

Putting this value of v in (i)

n²h²/4π²m²r² =GM/r

→r =n²h²/(4π²GMm²)

Now total energy of the system will be the sum of the kinetic energy and the potential energy. The kinetic energy,

K =½mv² 

  =½mGM/r 

  =½mGM*(4π²GMm²)/n²h²

  =2π²G²M²m³/n²h² 

The potential energy,

P =-GMm/r  

  =-GMm*(4π²GMm²/n²h²) 

  =-4π²G²M²m³/n²h² 

Hence the total energy of the system,

=K+P

=2π²G²M²m³/n²h² -4π²G²M²m³/n²h²

=-2π²G²M²m³/n²h².   





 

     44.  A uniform magnetic field B exists in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr's quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit, and (c) the minimum possible speed of the electron.  


ANSWER: (a) Force on the electron in the magnetic field =evB. 

For the circular motion of the electron, the centripetal force =mv²/r 

Equating the two, 

mv²/r =evB 

→r =mv/eB ------- (i)

From Bohr's quantiztion rule, 

mvr =nh/2π

→mv =nh/2πr         

evB=mv²/r; mv=eBr

From (i), 

r =nh/(2πreB) 

→r² =nh/(2πeB) 

For the smallest possible radius, n =1. Putting n =1 in the above expression,

r² =h/(2πeB) 

→r =√{h/(2πeB)} 


(b) Since r² =nh/(2πeB), the radius of the nth orbit will be, 

r =√{nh/(2πeB)}. 


(c) From (i), 

mv =erB 

→v =erB/m 

Putting the value of r, 

v =(eB/m)*√{nh/(2πeB)} 

  =√[nhe²B²/(2πm²eB)] 

 =√[nheB/(2πm²)] 

Except for n, all other entities are constant in this expression. Hence for the lowest possible speed of the electron n =1. It is equal to 

v =√[heB/(2πm²)].      





 

     45.  Suppose in an imaginary world the angular momentum is quantized to be an even integer multiple of h/2π. What is the longest possible wavelength emitted by hydrogen atoms in the visible range in such a world according to Bohr's model?  


ANSWER: The visible range of wavelength is from 365.0 nm to 656.3 nm. Since the visible range is obtained from the transition of electrons from the higher state to the n =2 state. Let us check the state for the longest wavelength of the visible range i.e. 656.3 nm. 

1/(656.3 nm) =1.097x10⁷{1/2² -1/n²} 

→1/4 -1/n² =1/(1.097x10⁷*656.3x10⁻⁹)

→1/4 -1/n²=0.139 

→1/n² =0.11 

→n² =9 

→n =3       

But in the given imaginary world the angular momentum is quantized to be an even integer multiple of h/2π. So the state n =3 is not possible. So we will take the next higher even multiple, i.e. n =4. The longest wavelength 𝜆 is given as

1/𝜆 =1.097x10⁷{1/2² -1/4²}

   =0.206x10⁷

→𝜆 =4.86x10⁻⁷ m

      =486x10⁻⁹ m

      =486 nm.

 



 

     46.  Consider an excited hydrogen atom in state n moving with a velocity v (v << c). It emits a photon in the direction of its motion and changes its state to a lower state m. Apply momentum and energy conservation principles to calculate the frequency 𝜈 of the emitted radiation. Compare this with frequency 𝜈ₒ emitted if the atom were at rest.  


ANSWER: Though the velocity of hydrogen atom v << c, the emitted photon will have the velocity =c because it is electromagnetic radiation. The momentum of the photon p =E/c. Where E is the energy of the photon.

From the momentum conservation principle, 

mv =mv' +p

→v' =v -p/m    

From the energy conservation principle,

½mv² =½mv'² +E

→v² =v'² +2E/m

→v'² =v² -2E/m

→v² -p²/m² -2vp/m =v² -2E/m

→2E/m =p²/m² +2vp/m

→2E/m  =E²/m²c² +2vE/mc

→E/mc² +2v/c =2

→E/mc² =2(1 -v/c)

→E =2mc²(1 -v/c) ------- (i)

 When the hydrogen atom is at rest. Let the energy of the emitted photon =E'. Its momentum p' =E'/c.

From the momentum conservation principle, 

mv' +p' =0

→mv' =-p'

→v' =-p'/m =-E'/mc

From the energy conservation principle, 

½mv'² =E'

→v'² =2E'/m

→E'²/m²c² =2E'/m

→E' =2mc²  --------- ----- (ii)


If 𝜈 is the frequency of the radiation when the hydrogen atom is moving and 𝜈ₒ the frequency of the radiation when the hydrogen atom is at rest then,

E =h𝜈 =2mc²(1 -v/c),  from (i)

And E' =h𝜈ₒ =2mc², from (ii)

Hence, 

E/E' =h𝜈/h𝜈ₒ =2mc²(1- v/c)/2mc²

→𝜈/𝜈ₒ =(1 -v/c)

→𝜈 =𝜈ₒ(1 -v/c).  

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Links to the Chapters





CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics




CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion









CHAPTER- 11 - Gravitation





CHAPTER- 10 - Rotational Mechanics






CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

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CHAPTER- 7 - Circular Motion

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For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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