Wednesday, May 3, 2023

H C Verma solutions, X-rays, Chapter-44, OBJECTIVE-I, Concepts of Physics, Part-II

X - rays

OBJECTIVE-I

      1.  X-ray beam can be deflected
(a) by an electric field
(b) by a magnetic field
(c) by an electric as well as a magnetic field
(d) neither by an electric field nor by a magnetic field.  

Answer: (d).    


Explanation:  X-rays are not charged particles, they are transverse electromagnetic radiation waves. These waves have a dual nature. Even when explained through quantum physics as being made of photons, these photons are also not electrically charged. Hence X-rays can not be deflected by an electric or magnetic field. Option (d) is correct.        







      2.  Consider a photon of continuous X-ray coming from a Coolidge tube. Its energy comes from 
(a) the kinetic energy of the striking electron 
(b) the kinetic energy of the free electrons of the target 
(c) the kinetic energy of the ions of the target
(d) an atomic transition in the target.  

Answer: (a).    


Explanation: When a striking electron enters the target, it collides with the atoms of the target before being stopped. During the collision, its kinetic energy is used to increase the temperature of the target or release X-ray photons. This is how continuous X-ray is produced. Option (a) is correct.         





      3.  The energy of a photon of a characteristic X-ray from a Coolidge tube comes from   
(a) the kinetic energy of the striking electrons 
(b) the kinetic energy of the free electrons of the target
(c) the kinetic energy of the ions of the target
(d) an atomic transition in the target.    
 

Answer: (d).    


Explanation: Characteristic X-rays are produced when the striking electrons dislodge inner orbit electrons and the vacancy thus produced in the inner orbit is filled by the transition from outer orbit electrons. The X-ray radiation thus emitted is called characteristic X-ray because its frequency is related to the material of the target. Option (d) is correct.      





      4.  If the potential difference applied to the tube is doubled and the separation between the filament and the target is also doubled the cutoff wavelength
(a) will remain unchanged 
(b) will be doubled
(c) will be halved
(d) will become four times the original. 

Answer: (c).    


Explanation: The cutoff wavelength is given as 
  𝝀ₘᵢₙ =hc/eV
Where h is the plank's constant, c is the speed of light and V is the potential difference across the Coolidge tube. 
    When V is doubled,
  𝝀'ₘᵢₙ =hc/(e*2V) =½𝝀ₘᵢₙ  
So the cutoff wavelength will be halved. 
Option (c) is correct. 

Note that it is not dependent on the separation between the filament and the target.         





      5.  If the current in the circuit for heating the filament is increased, the cutoff wavelength
(a) will increase
(b) will decrease
(c) will remain unchanged
(d) will change.  

Answer: (c).    


Explanation: The cutoff wavelength depends upon the potential difference V between the filament and the target, and not on the current in the filament. Since V is unchanged, the cutoff wavelength will remain unchanged. Option (c) is correct.  

        





      6.  Moseley's law for characteristic X-rays is √𝜈 =a(Z-b). In this 
(a) both a and b are independent of the material 
(b)  a is independent but b depends on the material 
(c) b is independent but a depends on the material 
(d) both a and b depend on the material. 

Answer: (a).    


Explanation: Both the constants a and b are independent of the material. The value of a =√(3Rc/4), all constants and the value of b depends upon the orbit from which the transition of an electron is about to take place. Option (a) is correct.   

      





      7.  Frequencies of Kₐ X-rays of different materials are measured. Which one of the graphs in Figure (44-Q1) may represent the relation between the frequency 𝜈 and the atomic number Z.
Figure for Q-7

 

Answer: (d).    


Explanation: From Moseley's law, 
  √𝜈 = a(Z-b)
→𝜈 =a²(Z -b)²

It is an equation of a parabola with some intercept on the Z-axis. 
Hence graph d shows the relation between frequency 𝜈 and atomic number Z. 
Option (d) is correct.


 





      8.  The X-ray beam coming from an X-ray tube 
(a) is monochromatic 
(b) has all wavelengths smaller than a certain maximum wavelength.
(c) has all wavelengths greater than a certain minimum wavelength 
(d) has all wavelengths lying between a minimum and a maximum wavelength.    

Answer: (c).    


Explanation: There is a certain minimum wavelength called cutoff wavelength for a given potential difference between the filament and the target below which no wavelength is found. All other wavelengths greater than the cutoff wavelength are present in the X-ray beam coming from an X-ray tube. Option (c) is correct.   


         





      9.  One of the following wavelengths is absent and the rest are present in the X-rays coming from a Coolidge tube. Which one is the absent wavelength? 
(a) 25 pm            (b) 50 pm 
(c) 75 pm            (d) 100 pm.   

Answer: (a).    


Explanation: The wavelengths shorter than the cutoff wavelength are not present in the X-rays coming from a Coolidge tube. If only one wavelength out of the given four wavelengths is absent in the coming X-ray radiation, it must be the shortest one that would be less than the cutoff wavelength. Option (a) is correct. 


           





      10.  Figure (44-Q2) shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for tube A in which the potential difference between the target and the filament is VA and the atomic number of the target material is ZA. These quantities are VB and ZB for the other tube. Then 
(a) VA > VB, ZA > ZB 
(b) VA > VB, ZA < ZB 
(c) VA < VB, ZA > ZB
(d) VA < VB, ZA < ZB.    
Figure for Q-10


Answer: (b).    


Explanation: Cutoff wavelength is inversely proportional to the potential difference between the target and the filament. The cutoff wavelength shown in the graph is shorter in tube A, hence the potential difference VA > VB.  
    The Kₐ wavelength in tube B is shorter than the Kₐ wavelength in tube A, which means the Kₐ frequency in tube B is greater than the Kₐ frequency in tube A. 
   From Moseley's law
 Frequency 𝜈 =a²(Z-b)²
Clearly, the Z of the target in tube B is greater than the Z in the target in tube A.
So, ZA < ZB. 
So option (b) is correct.

 





      11.  50% of the X-ray coming from a Coolidge tube is able to pass through a 0.1 mm thick aluminum foil. If the potential difference between the target and the filament is increased, the fraction of the X-ray passing through the same foil will be
(a) 0%             (b) <50% 
(c) 50%           (d) >50%.    

Answer: (d).    


Explanation: The penetrating power of an X-ray photon depends on its energy. When the potential difference between the target and the filament is increased, the cutoff wavelength is decreased which means that more energetic photons are now being produced. These increased energy photons will now fully pass through the given thickness of the aluminum foil. Thus more than 50% of the X-rays will be able to pass through the aluminum foil. Option (d) is correct.  



          





      12.  50% of the X-ray coming from a Coolidge tube is able to pass through a 0.1 mm thick aluminum foil. The potential difference between the target and the filament is increased. The thickness of aluminum foil, which will allow 50% of the X-ray to pass through, will be 
(a) zero               (b) <0.1 mm 
(b) 0.1 mm          (d) >0.1 mm.    

Answer: (d).    


Explanation: As we have seen in problem (11) that when the potential difference is increased in the Coolidge tube, more than 50% of the X-rays are able to pass through the 0.1 mm thick aluminum foil. It is due to the fact that more energetic X-rays are now being produced that all pass through the given foil. To limit the fraction of passing through X-rays to 50%, we need to restrict a part of these passing energetic  X-rays. For this, we will need a thicker foil. So Option (d) is correct.  


     





      13.  X-ray from a Coolidge tube is incident on a thin aluminum foil. The intensity of the X-ray transmitted by the foil is found to be Iₒ. The heating current is increased so as to increase the temperature of the filament. The intensity of the X-ray transmitted by the foil will be
(a) zero              (b) <Iₒ 
(c) Iₒ                  (d) >Iₒ. 

Answer: (d).    


Explanation: When the potential difference between the target and the filament is kept the same but the heating current is increased to increase the temperature of the filament, more electrons from the filament are available for being accelerated. The kinetic energy of all electrons reaching the target remains the same. Due to this, the intensity of the produced X-rays increases without increasing the penetrating power. Thus the same fraction of the X-rays will pass through the foil but with increased intensity. So Option (d) is correct.  


      





      14.  Visible light passing through a circular hole forms a diffraction disc of radius 0.1 mm on a screen. If an X-ray is passed through the same setup, the radius of the diffraction disc will be 
(a) zero             (b) <0.1 mm 
(c) 0.1 mm        (d) >0.1 mm.   

Answer:  (b).   


Explanation: The radius of a diffraction disc for a given hole and the distance of the screen are directly proportional to the wavelength of the passing light. Since the wavelength of X-rays is smaller than the wavelength of visible light, the radius of the diffraction disc for X-rays for the same setup will be smaller. Option (b) is correct.   

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Links to the Chapters


CHAPTER- 44- X-rays



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".

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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)

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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


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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"

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Click here for "OBJECTIVE-II"
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