Tuesday, January 3, 2023

H C Verma solutions, BOHR'S MODEL AND PHYSICS OF THE ATOM, Chapter-43, Questions for Short Answer, Concepts of Physics, Part-II

Bohr's Model and Physics of the Atom


Questions for Short Answer


     1.  How many wavelengths are emitted by atomic hydrogen in the visible range (380 nm - 780 nm)? In the range of 50 nm to 100 nm?  


ANSWER: The wavelengths emitted by atomic hydrogen are given as,

1/λ =R{1/n² -1/m²}, where R is the Rydberg constant.

In the Lyman series, the longest wavelength is for n =1 and m =2,

1/λ =R{1/1² -1/2²} =0.75R

→λ =1/(0.75*1.097x10⁷) m 

      =1.215x10⁻⁷ m

     =121.5 nm

So all other wavelengths in this series are <121.5 nm <380 nm.

For the Balmer series, the longest wavelength is for n =2 and m =3.

1/λ =R{1/2² -1/3²} =5R/36

→λ =36/(5*1.097x10⁷) m

      =6.56x10⁻⁷ m =656 nm.

Which is in the visible range.

The next wavelength in this series is for n =2 and m =4.

1/λ =R{1/2² -1/4²} =3R/16

→λ =16/(3*1.097x10⁷) m

    =4.86x10⁻⁷ m =486 nm.

This is also in the visible range. 

The next wavelength in this series is for n =2 and m =5. 

So 1/λ =R{1/2² -1/5²} =21R/100

→λ =100/(21x1.097x10⁷) m

      =4.34x10⁻⁷ m =434 nm.

It is also in the visible range.

Next for n =2 and m =6,

1/λ =R{1/2² -1/6²} =8R/36 =2R/9

→λ =9/(2*1.097x10⁷) m

      =4.10x10⁻⁷ m =410 nm.

It is also in the visible range.

Next for n =2 and m =7,

we get λ =(4x49)/45R

           =397 nm. 

It is also in the visible range. 

For n =2 and m =8, 

λ =64/15R =389 nm. It is more than 380 nm, so also in the visible range.

Next for n =2 and m =9,

λ = (81*4)/77R =384 nm, also in the visible range because it is more than 380 nm.

For n =2 and m =10,

   λ =100/24R =379.6 nm which is just less than 380 nm. So not in the range.

     Thus in the Balmer series, there are eight emitted wavelengths between 380 nm to 780 nm. 

      Let us check in the Paschen series. The shortest wavelength in this series is for n =3 and m =∞.  

So λ =9/R =820 nm. Which is more than 780 nm and not in the visible range. All other wavelengths in this series will be longer than it and not in the visible range.

   Thus a total of eight wavelengths are emitted by atomic hydrogen in the given visible range.


     As we have seen in the beginning, the longest wavelength in the Lyman series is 121.5 nm, the wavelengths between 50 nm to 100 nm will be found in this series only. Here n =1. Let us calculate for m =3. 

1/λ =R{1/1² -1/3²} =8R/9

→λ =9/{8*1.097x10⁷} m

     =1.025x10⁻⁷ m =102.5 nm 

For m =4, λ =16/15R =97.2 nm.

For m =5, λ =25/24R =95 nm.

For m =6, λ =36/35R =93.7 nm.

 and so on till m =∞, where λ =1/R

→λ =91.1 nm.

Thus we see that for m >4 to infinite there is an infinite number of closely spaced wavelengths between 50 nm to 100 nm. 





     2.  The first excited energy of a He⁺ ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom? 


ANSWER: The ground state energy of hydrogen is given as, 

E = -(13.6 eV)Z²/n² =-13.6 eV

{Since Z =1, n =1} 

For the first excited state energy of He⁺ ion, (Z =2, n =2), 

E' =-13.6 eV

E =E'.

We see that if only Z =n, (n-1)th excited state of the hydrogen-like ion will be equal to the ground state energy of the hydrogen ion. So it is not always true.  




     3.  Which wavelengths will be emitted by a sample of atomic hydrogen gas (in the ground state) if electrons of energy 12.2 eV collide with the atoms of the gas?  


ANSWER: In collisions, the hydrogen atom may absorb a part of the energy of the moving electrons to reach higher excited states. If we assume that after the collision the maximum energy of the colliding electron 12.2 eV is absorbed by the hydrogen atom then the total energy of the hydrogen atom now is, 

E =-13.6 eV +12.2 eV =-1.4 eV. 

Suppose with this energy the electron of the hydrogen atom jumps to the nth orbit. But the energy of the excited hydrogen atom with the electron in the nth orbit is given as, 

E =-13.6/n² eV. 

Equating, 

-1.4 =-13.6/n²

→n² =13.6/1.4 =9.71

→n =3.1  


   Since n can take only integer values, the colliding electrons can transfer only that much energy which is needed to take the atom to 1st or 2nd excited state, i.e. n =2 or 3.

        When the electron returns back to the ground state i.e. from n =3 to n =1, and n =2 to n =1, the two wavelengths of radiation emitted will be in the wavelength range of the Lyman series. 

Wavelength of radiation for n =1, m =3;

1/λ =R{1/1² -1/3²} 

      =8R/9

→λ =9/8R

      =9/(8*1.097x10⁷) m

      =1.025x10⁻⁷ m

      =102.5 nm.

Wavelength of radiation for n =1, m =2;

1/λ =R{1/1² -1/2²}

→λ =4/3R

     =4/(3*1.097x10⁷) m

     =1.215x10⁻⁷ m

     =121.5 nm. 






  

     4.  When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed in the Lyman series only. Explain?  


ANSWER: White radiation is X-ray radiation having photon energies of more than 41.4 eV. Hydrogen gas at room temperature is in the ground state i.e. n =1. When the white radiation is passed through the hydrogen gas at room temperature, the hydrogen atoms may absorb that much energy from the colliding photons which are required to take its electrons to higher energy levels, i.e. from n =1 to n =2, 3, 4, ....

     Hence in the absorption spectrum, wavelengths corresponding to these energy absorptions (λ =hc/E) are missing and such absorption lines are observed. These are the same wavelengths that are found in the emission phase when electrons jump back to the ground state (n =1) from a higher energy level. These wavelengths are named as Lyman series. Other series of wavelengths are not observed in the absorption spectra because energies are not being absorbed from excited states of hydrogen atoms i.e. from n =2, 3, ....





  


     5. The Balmer series was observed and analyzed before the other series. Can you suggest a reason for such an order?  


ANSWER: The Balmer series corresponds to radiations due to the jumping back of electrons from higher energy levels to n =2. These wavelengths range from 656.3 nm to 365 nm. Since most parts of the Balmer series fall in the visible range of 380 nm to 780 nm, it was natural to be observed and analyzed before other series.  




     6.  What will be the energy corresponding to the first excited state of hydrogen if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write Eₙ =E₁/n²? rₙ =aₒn²?   


ANSWER: Normally when the electron is widely separated from the proton, the potential energy of the atom is assumed to be zero. In this case, since the potential energy is already 10 eV, the energy of the first excited state of the hydrogen atom will be,

=-10 -13.6/2² eV 

=-10-3.4 eV 

=-13.4 eV. 

       

We can still write Eₙ =E₁/n² and rₙ =aₒn² because the effects of the reference point for potential energy have already been taken in calculating E₁ and aₒ.





   

     7.  The difference in the frequencies of the series limit of the Lyman series and the Balmer series is equal to the frequency of the first line of the Lyman series. Explain.  


ANSWER: The series limit is the highest frequency of that series. So for the series limit, the energy of the photons is maximum in that series. Electrons will release maximum energy if they jump down to the highest number of energy levels. 

  So the energy limit frequency in the Lyman series will be found when electrons jump down from n =∞ to n=1. And in the Balmer series from n =∞ to n =2.

   So energy limit of the Lyman series,

f =k{1/n² -1/m²}, where k is a constant.

  =k{1/1² -1/∞²} 

  =k.

   And the energy limit of the Balmer series, 

f' =k{1/2² -1/∞²}

   =k/4


Now, the first line of the Lyman series is the shortest frequency of the series corresponding to the jumping down of electrons from n =2 to n =1. This frequency is,

f" =k{1/1² -1/2²}

   =3k/4.

We see that the difference in the series limits of these two series is 

=k -k/4

=3k/4

=f".



     




     8.  The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units? 


ANSWER: The potential difference through which an electron should be accelerated to acquire that much energy that is required to completely detach it from the nucleus is called ionization potential. So the unit of ionization potential is Volts. Now the energy required to completely detach an electron from the nucleus is the ionization energy. When we express this ionization energy in eV, then by definition it is the energy required to accelerate an electron through a potential difference of V volts. So, numerically ionization energy and ionization potential are equal. 

   It will not be the case if the ionization energy is expressed in other units. 



    




     9.  We have stimulated emissions and spontaneous emissions. Do we also have stimulated absorption and spontaneous absorption?  


ANSWER: When an atom in a higher energy state is left there, it will eventually come down to a lower state by emitting a photon of energy that is equal to the difference in energy levels. This emission is called spontaneous emission. But when an atom emits a photon due to its interaction with a photon incident on it, the process is called stimulated emission. The emitted photon has exactly the same energy, phase, and direction as the incident photon. 

   When an atom in a lower energy state is incident upon by a photon having energy equal to the difference of energy levels, it may absorb the photon and jump to a higher energy state. This process is called stimulated absorption. However, there can not be spontaneous absorption because an atom can not jump to a higher energy state itself without absorbing a photon.  




      




     10.  An atom is in its excited state. Does the probability of its coming to the ground state depend on whether the radiation is already present or not? If yes does it also depend on the wavelength of the radiation present?  


ANSWER: An atom in its excited state does not remain so for long if it is left to itself. It eventually comes down to the ground state after the "lifetime of that state" by emitting a photon. But when radiation is present, the excited atom may interact with the photon and come down to the ground state by emitting a similar photon. It is called stimulated emission. So the probability of the atom coming down to the ground state depends upon the radiation present. 

      The energy of a photon depends upon the wavelength. In the stimulated emission the energy of the incident photon should be equal to the difference in the energy states because a similar photon is emitted after the interaction. So the coming down to the ground state in the presence of radiation also depends upon the wavelength of the radiation.     

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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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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)

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CHAPTER- 7 - Circular Motion

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CHAPTER- 6 - Friction

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Click here for → Friction - OBJECTIVE-II

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Click here for → EXERCISES (21-31)

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

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CHAPTER- 4 - The Forces

The Forces-

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Click here for "The Forces" - OBJECTIVE-I


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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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