Friday, March 31, 2023

H C Verma solutions, BOHR'S MODEL AND PHYSICS OF THE ATOM, Chapter-43, EXERCISES, Q21 TO Q30, Concepts of Physics, Part-II

Bohr's Model and Physics of the Atom


Exercises, Q21 to Q30


     21.  According to Maxwell's theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in the ground state if this rule is followed?     


ANSWER: We know that the angular momentum of the electron is, 

mvr =nh/2π.

Here n =1 and the radius of the orbit in this ground state r =0.053 nm. 

So, mvr =h/2π.   

Angular speed ⍵ =v/r

→⍵ =mvr/mr² =h/2πmr²

The frequency of revolution f is given as,

⍵ =2πf.

→f =⍵/2π

Since the frequency of radiation is equal to the frequency of revolution, f = c/λ, where λ is the wavelength and c is the speed of light.

Hence λ =c/f

         =2πc/⍵

      =2πc*2πmr²/h

      =4π²mr²c/h

=4π²*9.1x10⁻³¹*(0.053x10⁻⁹)²*3x10⁸/6.63x10⁻³⁴ m

=4.57x10⁻⁸ m

=45.7x10⁻⁹ m

=45.7 nm.

           






     22.  The average kinetic energy of molecules in a gas at temperature T is 1.5kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atom. Will hydrogen remain in its molecular form at this temperature? Take k = 8.62x10⁻⁵ eV/K.   


ANSWER: The binding energy of a hydrogen atom =13.6 eV.

The average kinetic energy of molecules of a gas at temperature T is given equal to 1.5kT.

From the given condition, 

1.5kT =13.6, putting the value of k from the given problem, 

→1.5*8.62x10⁻⁵*T =13.6 

→T =1.05x10⁵ K.  

  This is a very very high temperature and the hydrogen will not remain in its molecular form at this temperature.



  


 

     23.  Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n =3 state. Hydrogen can now emit red light of wavelength 653.1 nm.  Because of the Maxwellian distribution of speed, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.   


ANSWER: Energy required to take a hydrogen atom from ground state to n =3 state is, 

E =13.6{1/1² -1/3²} eV

  =13.6*8/9 eV 

  =12.09 eV. 

The average thermal kinetic energy is given as, 

     K.E. =1.5kT eV


      Equating both the energies as per the given condition,  

1.5kT =12.09 

→T =12.09/1.5k 

  =12.09/(1.5*8.62x10⁻⁵) K 

  =9.4x10⁴ K.                 





 

     24.  The average lifetime of a hydrogen atom excited to n = 2  state is 10⁻⁸ s. Find the number of revolutions made by the electron on average before it jumps to the ground state.     


ANSWER: Angular momentum of the electron of a hydrogen atom at n =2, 

mvr =nh/2π =h/π

Angular speed ⍵ =v/r =mvr/mr²

→⍵ =h/πmr²

Frequency f is given as,

⍵ =2πf

→f =⍵/2π

   =h/{2π²mr²}

  =h/{2π²m(n²aₒ)²}, here n =2, so

f =h/{32π²maₒ²}

=6.63x10⁻³⁴/{32π²*9.1x10⁻³¹*(0.053x10⁻⁹)²} Hz

=8.21x10¹⁴ Hz

Hence the number of revolutions made by the electron in the average lifetime of t =10⁻⁸ s is

=f*t

=8.21x10¹⁴*10⁻⁸ 

=8.21x10⁶.            





 

     25.  Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.    


ANSWER: Current in the orbit (ground state) due to a revolving electron, 

i =Q/t =e/{2πaₒ/v}

 =ev/{2πaₒ}

But v =Ze²/(2εₒhn)

Putting Z =1 and n =1, we have 

v =e²/(2εₒh)

Now, i =e³/{4πεₒhaₒ}

Hence the magnetic dipole moment,

 =i*A

 =(1/4πεₒ)*{e³/(haₒ)}*(πaₒ²)

=9x10⁹*(1.6x10⁻¹⁹)³*πaₒ/h

=1.16x10⁻⁴⁶*0.053x10⁻⁹/6.63x10⁻³⁴ A-m²

=9.2x10⁻²⁴ A-m².

     





 

     26.  Show that the ratio of the magnetic dipole moment to the angular momentum (l =mvr) is a universal constant for hydrogen-like atoms and ions. Find its value.    


ANSWER: As we have seen in the previous problem, the magnetic dipole moment is 

µ =i*A

 =(ev/2πr)*πr² 

=evr/2   

The angular momentum

l =mvr  

Hence the ratio,

µ/l =evr/(2mvr) 

   =e/2m =½(e/m) 

Since e/m is the ratio of charge to mass of an electron which is a universal constant, hence µ/l =½(e/m) is also a universal constant. Its value is 

=½*(1.6x10⁻¹⁹/9.1x10⁻³¹) C/kg 

=8.8x10¹⁰ C/kg.             





 

     27.  A beam of light having wavelengths distributed uniformly between 450 nm to 550 nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?    


ANSWER: The given range of wavelengths lies in the visible range. Hence photons having energies equal to the transition of electrons from n >2 to n =2 may be absorbed by the hydrogen gas and these absorbed light wavelengths will have the least intensity in the transmitted beam. 

The wavelength for transition n =3 to n =2,

1/λ =R{1/4 -1/9} =5R/36

→λ =656 nm

For the transition from n =4 to n=2,

1/λ =R{1/4 -1/16} =3R/16

→λ =486 nm

For the transition n =5 to n =2,

1/λ =R{1/4 -1/25} =21R/100

→λ =434 nm.

   So only wavelength λ =486 nm of the given light beam (450 nm to 550 nm) may be absorbed by the hydrogen and this wavelength will have the least intensity in the transmitted beam.          





 

     28.  Radiation coming from transitions n =2 to n =1 of hydrogen atoms fall on helium ions in n =1  and n =2 states. What are the possible transitions of helium ions as they absorb energy from the radiation?    


ANSWER: The energy of radiation coming from transitions n =2 to n =1 of hydrogen atoms is 

E =13.6*{1 -1/2²} =10.2 eV. 

The energy required by an electron in helium ion for the transition from n =1 to n =2 states, 

E' =Z²*13.6*{1 -1/4} eV 

   =40.8 eV. 

It is more than 10.2 eV. Hence the photons coming from the transition in hydrogen atoms will not be absorbed by the helium ions for the transition from n =1 to n =2 states. All other transitions in helium ion from n = 1 state to higher states will require more than 40.8 eV of energy hence the coming photon will not be absorbed. Now consider the electron in the n = 2 state of the helium ion. The energy required by it for transition to n =3 states are,

E" =Z²*13.6{1/4 -1/9} =4*1.89 eV 

   =7.6 eV. 

For transition from n =2 to n =4 states    E* =4*13.6{1/4 -1/16} =10.2 eV. 

For the next higher transitions energy required will be more than 10.2 eV.  

The coming photons have energy =10.2 eV hence they may be absorbed by the helium ions for transitions from n = 2 to n =3 states or from n =2 to n =4 states that require energy less than or equal to 10.2 eV.            





 


     29.  A hydrogen atom in the ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. Assuming that the entire photon energy is taken up by the electron, with what kinetic energy will the electron be ejected?    


ANSWER: The energy of the photon of  wavelength 50 nm is 

E = hc/λ

=(4.14x10⁻¹⁵ eV-s)*3x10⁸ m/50x10⁻⁹ m

=24.8 eV

An electron in the ground state of a hydrogen atom requires 13.6 eV of energy to widely separate it from the nucleus. If this photon is fully absorbed by the electron then the rest of the energy will be conserved as the kinetic energy of the electron. Hence the electron will be ejected with a kinetic energy of 24.8 -13.6 eV =11.2 eV.

 





 

     30.  A parallel beam of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in the ground state. (a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon moving in the same direction as the incident photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam? (b) A radiation detector is placed near the gas to detect radiation coming from perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.    

 

ANSWER: (a) The energy of photons in the parallel beam, 

E = hc/λ

(Since λ =100 nm)

→E =4.14x10⁻¹⁵*3x10⁸/100x10⁻⁹ eV

     =12.42 eV

Since it is not necessary that this incident photon will essentially give its energy wholly or partly, a portion of this parallel beam will pass through the hydrogen gas intact. So in the transmitted beam original 100 nm wavelength will be observed.  

Suppose an electron in the ground state absorbs energy to jump into the next state n =2, the absorbed energy 

E =13.6{1 -1/2²} =10.2 eV

As per the given condition, the rest of the energy appears as another photon. Its energy

=12.42 -10.2 =2.22 eV

The wavelength of this observed photon

λ =hc/E

 =4.14x10⁻¹⁵*3x10⁸/2.22 m

 =5.60x10⁻⁷ m

 =560x10⁻⁹ m

 =560 nm.


If the electron jumps to n =3 state from the ground state, the absorbed energy

E =13.6{1 -1/9} =12.1 eV

The energy of the residual photon

=12.42 -12.1 =0.32 eV

Hence the observed wavelength of the transmitted beam

λ =4.14x10⁻¹⁵*3x10⁸/0.32 m

  =38.80x10⁻⁷ m

  =3880x10⁻⁹ m

  =3880 nm.


If the electron tries to jump to the n =4 state from the ground state, the required energy for absorption

 =13.6{1 -1/16} =12.75 

but it is more than the incident photon. So it will not be absorbed.


(b) Since in the above-excited states the electrons have jumped to n =3 and n =2 states. They will ultimately revert back to the ground state either directly or in steps. So possible transitions are, from m =3 to n=1, n =2 to n =1 and n =3 to n =2.

The wavelengths observed in these transitions will be:-

For n =3 to n =1

1/λ =R{1 -1/9} =8R/9

→λ =9/8R =1.03x10⁻⁷ m

      =103x10⁻⁹ m

      =103 nm.


For n =2 to n =1 transition

1/𝜆 =R{1 -1/4} =3R/4

→𝜆 =4/3R =1.21x10⁻⁷ m

      =121x10⁻⁹ m

      =121 nm.


For n =3 to n =2 transition,

1/λ =R{1/4 -1/9} =5R/36

→λ =36/5R =6.56x10⁻⁷ m

     =656x10⁻⁹ m

     =656 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

Click here for → Questions for Short Answer

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

Click here for "Questions for short Answers"


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"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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