Wednesday, May 17, 2023

H C Verma solutions, X-rays, Chapter-44, EXERCISES, Q21 to Q27, Concepts of Physics, Part-II

X - rays


EXERCISES, Q11 to Q20

       21. Find the maximum potential difference that may be applied across an X-ray tube with a tungsten target without emitting any characteristic K or L X-ray. The energy levels of the tungsten atom with an electron knocked out are as follows:-   
The cell containing vacancy       K            L          M   
Energy in keV           69.5      11.3       2.3 

ANSWER:  For the given condition, the accelerating electron's maximum energy should be just less than the energy required for knocking out an L shell electron. From the given data, the energy required for knocking out an L shell electron from the tungsten atom is 11.3 keV. Hence the striking electron's maximum energy should be less than 11.3 keV which requires a maximum potential difference applied across the given X-ray tube of less than 11.3 kV.      





 
       22. The electric current in an X-ray tube (from the target to filament) operating at 40 kV is 10 mA. Assume that an average, of 1% of the total kinetic energy of the electrons hitting the target are converted into X-rays. (a) What is the total power emitted as X-rays and (b) how much heat is produced in the target every second?  

ANSWER:  The total electric current in the X-ray tube, i =10 mA =0.01 A. 
The potential difference is V =40 kV.
 Hence the total power being consumed is, 
P =i*V =(0.01 A)*(40 kV) 
    =400 W 

(a) Since 1% of the total kinetic energy of the electrons hitting the target is converted into X-rays, hence 1% of the total power consumed will be emitted as X-rays which is equal to
=1% of 400 W =4 W.   

(b) Rest of the power is converted into heat i.e. =400 -4 W =396 W. 
1 W of power means 1J of energy per second. Hence every second 396 J of energy is converted into heat in the target.      




 
       23. Heat at the rate of 200 W is produced in an X-ray tube operating at 20 kV. Find the current in the circuit. Assume that only a small fraction of the kinetic energy of electrons is converted into X-rays.   

ANSWER:  If the loss of kinetic energy of electrons in converted X-rays is neglected, the total 200 W of power is converted into heat. Operating potential difference V =20kV.  
Hence the current in the circuit, 
i =Power/P.D.
  =(200 W)/(20 kV)
  =200/20000 A 
  =0.01 A
  =10 mA.       




 
       24. Continuous X-rays are made to strike a tissue paper soaked with polluted water. The incoming X-rays excite the atoms of the sample by knocking out the electrons from the inner shells. Characteristic X-rays are subsequently emitted. The emitted X-rays are analyzed and the intensity is plotted against the wavelength (figure 44-E1). Assuming that only Kₐ intensities are detected, list the elements present in the sample from the plot. Use Mosley's equation
    𝜈 =(25x10¹⁴ Hz)(Z -1)².    
The figure for Q-24


ANSWER:  From the given Mosley's equation,
(25x10¹⁴ Hz)(Z-1)² =𝜈 =c/𝜆
→(Z-1)² =3x10⁸/(25x10¹⁴)𝜆
→Z =1 +(√3/5)x10⁻³/√𝜆)

For 𝜆 =78.9 pm
Z =1 +(√3/5)x10⁻³/√(78.9x10⁻¹²)
   =1 +0.039x10³
   =1 +39
   =40
Hence one of the elements is Zr with Z =40.

For 𝜆 =146 pm,
Z =1 +(√3/5)x10⁻³/√(146x10⁻¹²)
   =1 +0.029x10³
   =1 +29
   =30
Hence one of the elements is Zn with Z =30.

For 𝜆 =158 pm
Z =1 +(√3/5)x10⁻³/√(158x10⁻¹²) 
     =1 +0.028x10³
     =1 +28
     =29
So one of the elements is Cu with Z =29. 

For 𝜆 =198 pm
Z =1 +(√3/5)x10⁻³/√(198x10⁻¹²)
    =1 +0.025x10³
    =1 +25  
    =26 
So the remaining element is Fe with Z =26.  

Hence the list of elements present in the sample on the basis of the plot is Zr, Zn, Cu, and Fe.       




 
       25. A free atom of iron emits Kₐ X-rays of energy 6.4 keV. Calculate the recoil kinetic energy of the atom. Mass of an iron atom =9.3x10⁻²⁶ kg.

ANSWER:  Energy of an X-ray photon (given), 
E = 6.4 keV. 
The linear momentum of this photon, 
p =E/c. 
After the emission of the photon, the atom will have the same linear momentum but in opposite directions. It is due to the law of conservation of linear momentum. Suppose the recoil speed of the atom is v, then its momentum =mv. 
           Equating the above two,  
    mv =p =E/c 
Recoiled-kinetic energy of the atom, 
K =½mv² 
   =(mv)²/2m
   =p²/2m
   =E²/2mc²  
    =(6.4x10³)²*1.602x10⁻¹⁹/{2*9.3x10⁻²⁶*(3x10⁸)²} eV 
{Since the denominator will be in Joule, to convert it to eV the numerator is multiplied by 1.602x10⁻¹⁹}
   =3.9x10⁻⁴ eV.       




 
       26. The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength (1/λ) of the light falling on the cathode. The potential difference applied across an X-ray tube is linearly related to the inverse of the cutoff wavelength (1/λ) of the X-ray emitted. Show that the slopes of the lines in the two cases are equal and find their value. 

ANSWER:  Stopping potential is given as
Vₒ =(hc/e)(1/λ) -φ/e
Where φ is the work function.
Here the slope of the line is,
 m =hc/e

The relation between the potential difference applied across an X-ray tube and the cutoff wavelength λ is given as
 λ =hc/eV
→V =(hc/e)(1/λ)

    The slope of this line is also the same and 
=hc/e
=4.14x10⁻¹⁵*3x10⁸ V-m
=12.42x10⁻⁷ V-m
=1.242x10⁻⁶ V-m.
 




 
       27. Suppose a monochromatic X-ray beam of wavelength 100 pm is sent through Young's double slit and the interference pattern is observed on a photographic plate placed 40 cm away from the slit. What should be the separation between the slits so that the successive maxima on the screen are separated by a distance of 0.1 mm? 

ANSWER:  For Young's double-slit interference pattern equation is 
w =z𝜆/d
Where 𝜆 is the wavelength of the radiation, d is the separation of the slits, z is the distance of the screen from the slit and w is the spacing of the successive maxima.
Here given that 𝜆 =100 pm =1x10⁻¹⁰ m.
z =40 cm =0.40 m,
w =0.1 mm =1x10⁻⁴ m
and d =?
 So, d =z𝜆/w
  =0.40*1x10⁻¹⁰/1x10⁻⁴ m
  =0.4x10⁻⁶ m
  =4x10⁻⁷ m.     

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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-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

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Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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Click here for → Friction - OBJECTIVE-II

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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


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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-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


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Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

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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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