Sunday, March 26, 2023

H C Verma solutions, BOHR'S MODEL AND PHYSICS OF THE ATOM, Chapter-43, EXERCISES, Q11 TO Q20, Concepts of Physics, Part-II

Bohr's Model and Physics of the Atom


Exercises, Q11 to Q20


     11.  Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.  


ANSWER: Coulomb force will be maximum when the electron is in the nearest possible orbit and it is in the ground state for which n =1. 

    The radius of the orbit is given as, 

r =εₒh²n²/(πmZe²)

For the hydrogen atom Z =1 and here n =1, 

So, r =εₒh²/(πme²)

=8.85x10⁻¹²*(6.63x10⁻³⁴)²/{π9.1x10⁻³¹*(1.6x10⁻¹⁹)²} m

=5.31x10⁻¹¹ m

    There is only a proton in the nucleus of the hydrogen, the charge on both the electron and the proton is e with opposite in nature, hence the Coulomb force on the electron is,

F =(1/4πεₒr²)e²

  =9x10⁹*(1.6x10⁻¹⁹)²/(5.31x10⁻¹¹)² N

  =8.2x10⁻⁸ N.  






 

     12.  A hydrogen atom in a state having a binding energy of 0.85 eV makes the transition to a state with an excitation energy of 10.2 eV. (a) Identify the quantum numbers n of the upper and lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.  


ANSWER: (a) The energy in a state of a hydrogen atom is numerically equal to the binding energy in that state. The energy in a state n is given as, 

Eₙ =E₁/n², where E₁ is the energy in the ground state that is equal to -13.6 eV. Hence

Eₙ =-13.6/n².

Given binding energy =0.85 eV, the energy in this state, Eₙ =-0.85 eV. 

Hence, 

n² =-13.6/Eₙ =-13.6/-0.85 =16

→n =4.


Given that for the final state, excitation energy =10.2 eV. Let the quantum number of this state =m. So, the energy of this state =-13.6 +10.2 eV =-3.4 eV.

Thus, m² =-13.6/-3.4 =4

→m =2.

So in this transition, the upper energy state is 4 and the lower energy state is 2.


(b) The wavelength 𝜆 of the emitted radiation is given as,

1/𝜆 =R{1/n² -1/m²}

Where m > n, So here n =2 and m =4

→1/𝜆 =1.097x10⁷{1/2² -1/4²}

→1/𝜆 =(3/16)*1.097x10⁷ m

→𝜆 =486x10⁻⁹ m =486 nm.

 





 

     13.  Whenever a photon is emitted by hydrogen in the Balmer series, it is followed by another photon in the Lyman series. What wavelength does this later photon correspond to?    


ANSWER: Since the photon is emitted in the Balmer series, the electron has jumped to the second energy level n =2. The following photon is emitted in the Lyman series, which means the electron here has jumped from n =2 to n =1. So the wavelength corresponding to the later emitted is given as 

  1/𝜆 =R{1/1² -1/2²} =3R/4

→𝜆 =4/3R

   =4/{3*1.097x10⁷} m

   =1.22x10⁻⁷ m

   =122x10⁻⁹ m

   =122 nm.

     





 

     14.  A hydrogen atom in state n =6 makes two successive transitions and reaches the ground state. In the first transition, a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition. (b) what is the value of n in the intermediate state?


ANSWER: The energy of the hydrogen atom in state n =6 is

E₆ =-13.6/n² =-13.6/6² eV

    =-0.38 eV.

Ground state energy, E₁ =-13.6 eV.

So the energy of intermediate state n is,

Eₙ =-0.38 -1.13 eV 

   =-1.51 eV


(a) So the energy of the photon emitted in the second transition

=Eₙ -E₁

=-1.51 -(-13.6) eV

=12.1 eV.


(b) Since Eₙ =-13.6/n²

→n² =-13.6/(-1.51) =9

→n =3.

 




 


     15.  What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?  


ANSWER: The kinetic energy of the electron in the nth orbit is, 

K =mZ²e⁴/(8εₒ²h²n²) 

In the ground state n =1, and for hydrogen Z =1. So the kinetic energy of the electron in the ground state,

K =me⁴/(8εₒ²h²) 

   =13.6 eV, when we put the respective values. 

Given that the potential energy in the ground state, V =0. 

So the total energy of the hydrogen atom in the ground state,

E₁ =K +V =13.6 eV. ------- (i)

   In the first excited state, n =2. We know that the energy of state n =2 is higher by 10.2 eV than the state n =1. {When we take the P.E. zero when the nucleus and electron are widely apart, E₁ =-13.6 eV and E₂ =-3.4 eV. So E₂ -E₁ =-3.4 -(-13.6) eV =10.2 eV}

    Hence the energy of the hydrogen atom in the first excited state, {from -(i)} 

=E₁ +10.2 eV

=13.6 +10.2 eV

=23.8 eV.             





 

     16.  A hot gas emits radiation of wavelength 46.0 nm, 82.8 nm, and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.  


ANSWER: Enengy corresponds to the photon of wavelength 46.0 nm

 =hc/𝜆 

 =6.63x10⁻³⁴*3x10⁸/46x10⁻⁹ J

 =4.32x10⁻¹⁸ J

 =27 eV.

The energy corresponding to the photon of wavelength 82.8 nm 

 =27*(46/82.8) eV

 =15 eV

The energy corresponding to the photon of wavelength 103.5 nm

 =27*(46/103.5) eV

 =12 eV.

 (i) Since the atom has only two excitation states, the photon corresponding to the radiation with energy 27 eV is for the transition of an electron from n =3 to n =1. Thus there is a difference of 27 eV of energy between n =1 and n=3. See the figure below,
The figure for Q-16

  (ii) Given that the difference between energy levels decreases as energy is increased. So the difference in energy between n =1 and n =2 is 15 eV and the difference of energy between n =2 and n =3 is 12 eV.

   If we take the energy of n =3 equal to zero then the energy of n =1 state (ground state) is

  =0 -27 eV =-27 eV.

{Refer para (i) above}


Similarly, the energy of n =2 state (First excitation state) is

 =0 -12 eV

 =-12 eV.

{Refer para (ii) above}   





 

     17.  A gas of hydrogen-like ions is prepared in a particular excited state A. It emits photons having a wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.  


ANSWER: Since there are six different wavelengths of emitted photons, there will be six different transitions. If an atom is excited to the principal quantum number n =m state then there will be m(m-1)/2 number of possible transitions. So here, 

m(m-1)/2 =6 

→m =4. 

Hence the principal quantum number of state A is 4.         

     If Z is the atomic number of the hydrogen-like ion, the energy of the nth state,

   E =-13.6Z²/n²

If there is a transition from state m to n where m > n, then the energy of the emitted photon

→E =-13.6Z²/m²-(-13.6Z²/n²) eV

      =13.6Z²{1/n²-1/m²} eV ---- (i)   

 

   Given that the wavelength related to one group of photons is equal to the first line of the Lyman series. The first line of the Lyman series is obtained when there is a transition of an electron from state n =2 to n =1 in a hydrogen atom. It has an energy of 

13.6{1-1/2²} eV =3*3.4 eV =10.2 eV. 

Equating energy in (i) to it we get,

 13.6Z²(1/n² -1/m²) =10.2 

→Z² =0.75m²n²/(m²-n²)


   Since atomic number Z will be a whole number, we try six different transition combinations of (m,n) to check what value is a whole number. The combinations will be (4,1), (4,2), (4,3), (3,1), (3,2), and (2,1).

     We find that two combinations (4,2) and (2,1) give whole numbers for Z.

For (4,2)→Z² =0.75*64/12 =4 

  →Z =2.

Which is for a Helium ion. 

For (2,1) →Z² =0.75*4/3 =1 

  →Z =1.

Which is for the hydrogen gas and the transition relates to the first line of the Lyme series itself. Since in the given problem, the gas in question is a hydrogen-like ion. Thus we take Z =2 for which the gas is identified as He⁺.

 





 

     18.  Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.  


ANSWER: The angular speed of the electron in a hydrogen atom is 

⍵ =v/r

Since angular momentum is

mvr =nh/2π

→v =nh/(2πmr)

  Putting this value of v in angular speed, we get

⍵ =nh/(2πmr²)    --------------- (i)

We see that the angular speed is directly proportional to 'n' but inversely proportional to the square of the radius. Thus the maximum angular speed of the electron of a hydrogen atom will be for n =1 where the value of radius is minimum i.e. r =0.053 nm. Putting the values in (i) we get,   

⍵ₘₐₓ=6.63x10⁻³⁴/{2π*9.1x10⁻³¹*(0.053x10⁻⁹)²} rads/s

=4.1x10¹⁶ rads/s





 

     19.  A spectroscope instrument can resolve two nearby wavelengths 𝜆 and 𝜆+Δ𝜆 if 𝜆/Δ𝜆 is smaller than 8000. This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument?  

Note: This problem is also explained in a video, Click this link to view it. 


ANSWER: Given 𝜆/Δ𝜆 < 8000.

For the Balmer series, the wavelength is given as

 1/𝜆 =R{1/4 -1/m²}

 Taking the differential on both sides, we get

-(1/𝜆²)*Δ𝜆 =(R*2/m³)Δm

→-(1/𝜆)*(Δ𝜆/𝜆)=2R*Δm/m³ 

→-R{1/4-1/m²}*(Δλ/λ) =2R*Δm/m³

→-(m²-4)(Δλ/λ)/4m² =2Δm/m³

The negative sign is for the fact that when m increases λ decreases. So we neglect it here. Also for the consecutive orbits, we take Δm =1. So now

(m²-4)m/8 =(λ/Δλ) <8000


By trial, for m =40, LHS =7980<8000

For m =41, LHS=8595 >8000

So this inequality is satisfied for m =40.

In this case of the Balmer series, there will be 38 transitions from m =40 to n=2 and only these 38 spectral lines will be resolved by the instrument.       

 

 

 


 

     20.  Suppose, in certain conditions, only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelengths emitted by hydrogen in the visible range (380 nm to 780 nm).  


ANSWER: (a) Smallest wavelength means the greatest energy. The energy of an orbit is inversely proportional to  n². So higher value of n, the curve of energy E becomes very close and asymptotic to the n-axis. Therefore the energy of a transition is more in orbits near the nucleus for the same difference of n. Therefore here we shall consider the transition of an electron from n =3 to n =1. Hence this wavelength will be given as, 

1/𝜆 =R{1/1² -1/3²} =8R/9

→𝜆 =9/8R

     =9/{8*1.097x10⁷}

     =1.03x10⁻⁷ m

     =103x10⁻⁹ m.

     =103 nm.


(b) The visible range lies in the Balmer series for which all the transitions are up to n =2. So in the given case, only the transition from n =4 to n =2 is considerable. The wavelength is 

1/𝜆 =R{1/2² -1/4²}

   =3R/16

→𝜆 =16/3R

   =16/{3*1.097x10⁷} m

   =4.86x10⁻⁷ m 

   =486x10⁻⁹ m

   =486 nm. 

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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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

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

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

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Click here for "OBJECTIVE-II"

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