Monday, March 20, 2023

H C Verma solutions, BOHR'S MODEL AND PHYSICS OF THE ATOM, Chapter-43, EXERCISES, Q1 TO Q10, Concepts of Physics, Part-II

Bohr's Model and Physics of the Atom


Exercises, Q1 to Q10


     1.  The Bohr radius is given by aₒ =εₒh²/πme². Verify that the RHS has dimensions of length. 


ANSWER: Dimensions of the entities on RHS are as follows:-

Dimensions of εₒ =[M⁻¹L⁻³T⁴I²]

Dimensions of h =[ML²T⁻¹]

π is dimensionless

Dimension p is m =[M]

Dimensions of e =[IT]

Hence the dimensions of RHS are,

[M⁻¹L⁻³T⁴I²][ML²T⁻¹]²/[M][I²T²]

=[MLT²I²]/[MI²T²]

=[L] =Dimension of length. 







     2.  Find the wavelength of the radiation emitted by hydrogen in the transitions (a) n =3 to n =2, (b) n =5 to n =4 and (c) n =10 to n =9. 


ANSWER: (a) The wavelength is given as

1/𝜆 =R{1/n² -1/m²}

Here n =2 and m =3,

So 1/𝜆 =R{1/4 -1/9} =5R/36

         =5*1.097x10⁷/36

→𝜆 =656x10⁻⁹ m =656 nm.


(b) Here, n =4, m =5, hence

1/𝜆 =R{1/16 -1/25} =9R/(16*25)

→𝜆 =400/9*1.097x10⁷ m

     =4050x10⁻⁹ m

     =4050 nm.


(c) Here n =9 and m =10. 

So 1/𝜆 =R{1/81 -1/100}

          =19R/8100

→𝜆 =8100/19*1.097x10⁷ m

      =38862x10⁻⁹ m

      =38862 nm.

 




 

     3.  Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He⁺, and (c) Li⁺⁺. 


ANSWER: The smallest wavelength will be when the energy released by the down-jumping electron is maximum. And the energy dissipated will be maximum when the electron jumps from n = infinity to n =1, i.e. to the ground state. For hydrogen or hydrogen-like atoms, the wavelength of the emitted radiation is given as

 1/𝜆 =RZ²{1/n² -1/m²}

where R is Rydberg's constant and Z is the atomic number.


(a) For hydrogen, Z =1, n =1 and m =∞. Hence the smallest wavelength is given as

1/𝜆 =R{1/1² -1/∞²} =R

→𝜆 =1/R

      =1/1.097x10⁷ m

      =91x10⁻⁹ m

      =91 nm.


(b) For He⁺, Z =2. Hence the smallest wavelength is given as

1/𝜆 =R*2²{1/1² -1/∞²} =4R

→𝜆 =1/4R

     =1/4*1.097x10⁷ m

     =23x10⁻⁹ m

     =23 nm.


(c) For Li⁺⁺ Z =3. Hence the smallest wavelength is given as,

1/𝜆 =R*3²{1/1² -1/∞²}

      =9R

→𝜆 =1/9*1.097x10⁷ m

     =10x10⁻⁹ m

     =10 nm.

 

 




 

     4.  Evaluate the Rydberg constant by putting the values of the fundamental constants in its expression. 


ANSWER: Expression for Rydberg constant R is given as

R =me⁴/8εₒ²h³c

Value of εₒ =8.85x10⁻¹² C²-m/N

mass of an electron, m =9.1x10⁻³¹ kg

Charge on an electron, e =1.6x10⁻¹⁹ C

Planks constant h =6.63x10⁻³⁴ J-s

Speed of light c =3x10⁸ m/s

Hence,

R =9.1x10⁻³¹*(1.6x10⁻¹⁹)⁴/8*(8.85x10⁻¹²)²*(6.63x10⁻³⁴)³*3x10⁸ m⁻¹

 ≈1.09x10⁷ m⁻¹.

 




 

     5.  Find the binding energy of a hydrogen atom in the state n =2. 


ANSWER: The binding energy of an atom in a given state is the energy needed to separate its constituents over large distances. Since the energy of a hydrogen atom for a state n is

E =-13.6/n² eV, it will require energy equal to 13.6/n² eV to separate the electron from the nucleus over a large distance. Hence the binding energy for the hydrogen atom in the state n =2 is

E' =13.6/2² eV =3·4 eV.       






 

     6.  Find the radius and energy of a He⁺ ion in the states (a) n =1, (b) n =4, and (c) n =10. 


ANSWER: The radius and energy of a hydrogen-like ion are given as,

rₙ =n²aₒ/Z, and Eₙ =-13.6Z²/n²

For He⁺ ion, Z =2,

Hence, rₙ =½n²aₒ, and Eₙ =-54.4/n² eV.


(a) For n =1, radius

r₁ =½*1²*aₒ 

    =½*53x10⁻¹² m

    =0.265x10⁻¹⁰ m

    =0.265 Å.

And the energy,

E =-54.4/1² eV

   =-54.4 eV.


(b) For n =4. 

Radius r =½*4²*aₒ

  →r =8*53x10⁻¹² m

       =4.24x10⁻¹⁰ m

       =4.24 Å.

Energy, E =-54.4/4² eV

               =-3.4 eV. 


(c) For n =10, 

Radius r =½*10²*53x10⁻¹² m

            =26.5x10⁻¹⁰ m

            =26.5 Å.

Energy E =-54.4/10² eV

             =-0.544 eV.

  




 

     7.  A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition? 


ANSWER: Since the emitted wavelength is in the ultraviolet region of the spectrum, it falls under the Lyman series. Here the electrons jump down to the ground state, n =1. Given that λ =102.5 nm.

 Rydberg's constant R =1.097x10⁷ m⁻¹

Let us assume that the transition of the electron is from quantum number n =m to n =1. Hence,

 1/λ =R{1/1² -1/m²}

→1/102.5x10⁻⁹ =1.097x10⁷(1-1/m²)

→1 -1/m² =0.889

→1/m² =1-0.889 =0.111

→ m² =9

→m =3.

So the quantum numbers of the states involved in the transition are 1 and 3.        




 

     8.  (a) Find the first excitation potential of He⁺ ion. (b) Find the ionization potential of Li⁺⁺ ion.  


ANSWER: (a) The excitation energy for the He⁺ ion for the first excited state is

E =-13.6Z²{1/1² -1/2²} eV

   =-10.2*Z² eV

   =-10.2*2² eV    {Here Z =2}

   =-40.8 eV. 

Hence the first ionization potential of the He⁺ ion is 40.8 V.


(b) For Li⁺⁺ ion, Z =3.      

Ionization energy here is the energy needed to widely separate the electron from the nucleus which is equal to

   =-13.6*Z²  

   =-13.6*3² eV 

  =-122.4 eV.

Hence the ionization potential is 122.4 V.  





 

     9.  A group of hydrogen atoms is prepared in n =4 states. List the wavelengths that are emitted as the atoms make transitions and return to n =2 states.  


ANSWER: Three types of wavelengths will be emitted for the transitions from 

(a) n =4 to n =2

(b) n =4 to n =3

then (c) n =3 to n =2. 

Wavelength for transition from n =4 to n =2 will be given as, 

1/𝜆 =R{1/2² -1/4²}

     =1.097x10⁷*3/16

     =2.057x10⁶

→𝜆 =487x10⁻⁹ m

     =487 nm. 


The wavelength for the transition from n =4 to n =3 will be given as 

1/𝜆 =R{1/3² -1/4²}

     =1.097x10⁷*7/(9*16) m

     =5.33x10⁵ m

→𝜆 =1876x10⁻⁹ m

     =1876 nm.


The wavelength for the transition from n =3 to n =2 will be given as 

1/𝜆 =R{1/2² -1/3²}

     =1.097x10⁷*5/36 

     =1.524x10⁶ 

→𝜆 =656x10⁻⁹ m

      =656 nm.

 




 

     10.  A positive ion having just one electron ejects it if a photon of wavelength 228 Å or less is absorbed by it. Identify the ion. 


ANSWER: Since the ion ejects the electron after absorbing the photon, the ionization energy of the ion should be equal to the energy of the photon.

𝜆 =228 Å =228x10⁻¹⁰ m

The energy of the photon,

E = hc/𝜆 

   =6.63x10⁻³⁴*3x10⁸/228x10⁻¹⁰ J

   =8.72x10⁻¹⁸ J. 

The ionization energy of an ion is given as

E' =13.6Z² eV, 

{where Z is the atomic number of the ion.} 

→E' =13.6x1.6x10⁻¹⁹Z² J.

Equating both the energies to solve for Z we get 

Z² =8.72x10⁻¹⁸/(13.6*1.6x10⁻¹⁹) 

→Z² =4 

→Z =2. 

Hence the ion, in this case, is He⁺.     

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

 Buy Home Furnishing

Click here for all links → kktutor.blogspot.com 

===<<<O>>>===


My Channel on YouTube  →  SimplePhysics with KK


Links to the Chapters


CHAPTER- 43- Bohr's Model and Physics of Atom



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"




Posted by at
Labels: , , , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)