Tuesday, May 9, 2023

H C Verma solutions, X-rays, Chapter-44, EXERCISES, Q1 to Q10, Concepts of Physics, Part-II

X - rays


EXERCISES, Q1 to Q10

      1.  Find the energy, frequency, and momentum of an X-ray photon of wavelength 0.10 nm.   

ANSWER:  Given that wavelength λ =0.10 nm.
So energy,
E =hc/λ
   =4.14x10⁻¹⁵*3x10⁸/0.10x10⁻⁹ 
   =124x10² eV
  =12.4 keV.

Frequency of the photon 
𝜈 =c/λ
   =(3x10⁸ m/s)/(0.10x10⁻⁹ m)
   =3x10¹⁸ Hz.

The momentum of the X-ray photon
p =E/c
   =(12.4x10³ eV)*(1.6x10⁻¹⁹ C)/(3x10⁸ m/s)
   =6.61x10⁻²⁴ kg-m/s.
 






 
      2.  Iron emits Kₐ X-ray of energy 6.4 keV and calcium emits Kₐ X-ray of energy 3.69 keV. Calculate the times taken by an iron Kₐ photon and a calcium Kₐ photon to cross through a distance of 3 km. 

Answer: Whatever the energy of an X-ray photon, they are all electromagnetic radiations and hence they all travel with the speed of light c. So here both the Kₐ photons will take the same time equal to
(3 km)/(3x10⁸ km/s) 
=1.0x10⁻⁸ s
=10 µs.   




 



      3.  Find the cutoff wavelength for the continuous X-rays coming from the X-ray tube operating at 30 kV.  

ANSWER:  The potential difference in the X-ray tube, 
V =30 kV. 
The cutoff wavelength is given as, 
λₘᵢₙ = hc/eV 
      =4.14x10⁻¹⁵*3x10⁸/(30x10³) m 
      =4.14x10⁻¹¹ m 
      =41.4 pm.   








      4.  What potential difference should be applied across an X-ray tube to get an X-ray of a wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in Joule?   

ANSWER:  Given λₘᵢₙ =0.10 nm.
Since λₘᵢₙ =hc/eV, where V =potential difference
→V =hc/λₘᵢₙ for h in eV-s 
       =4.14x10⁻¹⁵*3x10⁸/0.10x10⁻⁹ volts 
       =12.4x10³ volts 
       =12.4 kV.    
The maximum energy of a photon in this X-ray will be equal to the kinetic energy of the accelerated electron in the X-ray tube when all of the kinetic energy is converted into an X-ray photon. 
This energy E =eV 
→E=1.6x10⁻¹⁹x12.4x10³ J 
      ≈2x10⁻¹⁵ J.  



   




  
      5.  The X-ray coming from a Coolidge tube has a cutoff wavelength of 80 pm. Find the kinetic energy of the electrons hitting the target.  

ANSWER:  Given 𝜆ₘᵢₙ =80 pm.
But 𝜆ₘᵢₙ =hc/eV
Where eV is the kinetic energy of the electrons hitting the target.
→K.E. of electrons, eV =hc/𝜆ₘᵢₙ
  =4.14x10⁻¹⁵*3x10⁸/(80x10⁻¹²) eV
  =1.55x10⁴ eV
  =15.5 keV.
 




 


 
      6.  If the operating potential in an X-ray tube is increased by 1% by what percentage does the cutoff wavelength decrease?  

ANSWER:  Cutoff wavelength is given as 
λₘᵢₙ =hc/eV
When V is increased by 1%, new V'=1.01V
So now λ'ₘᵢₙ =hc/(1.01eV)
                   =0.99hc/eV  {Approx}
                   =0.99λₘᵢₙ     {Approx)
So the cutoff wavelength decreases by approximately 1%.  




 


 
      7.  The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30pm, find the electric field between the cathode and the target.   

ANSWER:  Given λₘᵢₙ =30 pm
Hence from λₘᵢₙ =hc/eV
→30x10⁻¹² =4.14x10⁻¹⁵*3x10⁸/eV
→V =41.4 kV
Distance between the filament and the target, d =1.5 m, hence the field between the two is
E =V/m =41.4x10³/1.5 =27.6 kV/m.
 




 



      8.  The short wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?   

ANSWER:  Let the original value of the operating voltage =V. New operating voltage,
V' =1.5V.  
If the original short wavelength limit =λ pm, then
λ*10⁻¹² =hc/eV
New short wavelength limit λ' =λ -26 pm
So, (λ -26)*10⁻¹² =hc/1.5eV 
→hc/eV -26x10⁻¹² =hc/1.5eV
→(hc/eV)*(1 -1/1.5) =26x10⁻¹²  
→hc/eV =78x10⁻¹²    
→V =4.14x10⁻¹⁵*3x10⁸/78x10⁻¹²
       =1.59x10⁴ volts
       =15.9 kV.  






 
      9.  The electron beam in a color TV is accelerated through 32 kV and then strikes the screen. What is the wavelength of the most energetic X-ray photon?   

ANSWER:  Given V =32 kV.
The wavelength of the most energetic X-ray photon is the cutoff wavelength, that is equal to hc/eV.  
Hence the required wavelength, 
  λₘᵢₙ  =hc/eV 
    =4.14x10⁻¹⁵*3x10⁸/32x10³ m 
    =0.388x10⁻¹⁰ m
    =38.8x10⁻¹² m 
    =38.8 pm.     






 
      10.  When 40 kV is applied across an X-ray tube, an X-ray is obtained with a maximum frequency of 9.7x10¹⁸ Hz. Calculate the value of the Plank constant from these data.   

ANSWER:  Given 𝜈ₘₐₓ =9.7x10¹⁸ Hz.
→𝜆ₘᵢₙ =c/𝜈ₘₐₓ
        =3x10⁸/9.7x10¹⁸ m
        =3.09x10⁻¹¹ m
Also given V =40 kV
But 𝜆ₘᵢₙ =hc/eV
→h = 𝜆ₘᵢₙ(eV)/c 
      =3.09x10⁻¹¹*40x10³/3x10⁸ eV-s
      =4.12x10⁻¹⁵ eV-s. 

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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 → 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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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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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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