Sunday, May 14, 2023

H C Verma solutions, X-rays, Chapter-44, EXERCISES, Q11 to Q20, Concepts of Physics, Part-II

X - rays


EXERCISES, Q11 to Q20

       11. An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.           

ANSWER:  After each collision, the photon energy will get reduced and a longer wavelength will be produced. Hence for the lowest three wavelengths emitted, we will consider the first three collisions.  
The potential difference in the X-ray tube =40 kV. 
The energy of the electron just before hitting the target =40 keV.
The energy of the first emitted photon,
E =70% of 40 keV =28 keV. 
The wavelength of this emitted radiation 
𝜆 =hc/E
   =4.14x10⁻¹⁵*3x10⁸/E
   =1242x10⁻⁹/E
   =1242/E nm
   =1242/28x10³ nm
   =0.0443 nm
   =44.3 pm.

The remaining energy of the electron =40 keV -28 keV =12 keV.
The energy of the photon emitted after the next collision =70% of 12 keV.
             =8.4 keV
The wavelength of this emitted radiation,
𝜆 =1242/8.4x10³ nm
   =0.148 nm
   =148 pm.

Now the remaining energy of the electron =12 keV -8.4 keV =3.6 keV.
The energy of the photon emitted after the next collision =70% of 3.6 keV =2.52 keV.
   The wavelength of this emitted radiation 
𝜆 =1242/2.52x10³ nm
   =0.493 nm
   =493 pm.
 
Further collisions will emit photons with even lower energies. Hence the emitted wavelengths will be longer.
 





 

       12. The wavelength of Kₐ X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having a tungsten target that allows the production of Kₐ X-ray?  

ANSWER: The energy difference between the K shell and the L shell will be equal to the energy of a Kₐ X-ray photon because this photon is emitted when an electron from the L shell fills a vacancy in the K shell of the atom. Here, the energy of a Kₐ X-ray photon,
E =hc/𝜆
   =1242/(21.3/1000) eV,
 {here wavelength should be in nm}
→E =58.3 keV.

For the production of K alfa X-ray, an electron in the K shell should be knocked out. Since the energy required to knock out an L shell electron is 11.3 keV, the energy required to knock out a K shell electron will be =11.3+58.3 keV =69.6 keV. 
   So an accelerated electron should have 69.6 keV energy before striking the target for the production of Kₐ X-ray. For an electron to be accelerated to 69.6 keV of energy, the accelerating voltage should be equal to 69.6 V.
 




 

       13. The Kᵦ X-ray of argon has a wavelength of 0.36 nm. The minimum energy needed to ionize an argon atom is 16 eV. Find the energy needed to knock out an electron from the K shell of an argon atom.             

ANSWER: An argon atom has electrons available up to the M shell. Hence the minimum energy to ionize an argon atom will be used to knock out an electron from the outermost M shell. 
     Kᵦ X-ray is produced when an electron makes a transition from the M shell to a vacancy in the K shell releasing a photon. Hence the energy of a photon of a Kᵦ X-ray will be equal to the difference in energy levels of a K shell and an M shell. 
   Here given wavelength of Kᵦ X-ray =0.36 nm. Hence the energy of a photon of this X-ray
E =hc/𝜆
   =1242/0.36 eV
   =3450 eV.

So the energy needed to knock out an electron from the K shell of an argon atom will be
=3450 ev +16 eV
=3466 eV
≈3.47 keV.
 




 

       14. The Kₐ X-rays of aluminum (Z =13) and Zinc (Z =30) have wavelengths 887 pm and 146 pm respectively. Use Mosley's law √𝜈 =a(Z-b) to find the wavelength of the Kₐ X-ray of iron (Z =26).           

ANSWER: 𝜈 =c/𝜆  =3x10⁸/𝜆*10⁻¹² Hz
                          {for 𝜆 in pm}
  →𝜈 =3x10²⁰/𝜆
  →√𝜈 =√(3/𝜆) *10¹⁰

Using Moseley's law for the given data for aluminum and zinc, we have
√(3/887) *10¹⁰ =a(13 -b)
→a(13 -b) =5.82x10⁸   ------- (i)

and
√(3/146)x10¹⁰ =a(30 -b)
→a(30-b)  =14.33x10⁸ ------- (ii)

Dividing (ii) by (i) we get
(30 -b)/(13 -b) =2.46
→30 -b =13*2.46 -2.46b
→1.46b =31.98 -30 =1.98
→b =1.356

From (ii)
a =14.33x10⁸/(30 -1.356)
   =5x10⁷

Hence the wavelength of Kₐ X-ray of Iron, (Z =26) will be given as
√(3/𝜆) *10¹⁰ =a(Z -b)
  →√(3/𝜆) *10¹⁰ =5x10⁷*(26-1.356)
 →3/𝜆 ={24.644*5/10³}² 
          =0.0152 
→𝜆 =3/0.0152 =197.4 pm.            




 

       15. A certain element emits Kₐ X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.          

ANSWER: E =3.69 keV. 
Hence the frequency of X-ray radiation, 
𝜈 =E/h
   =3.69x10³/4.14x10⁻¹⁵ Hz
   =8.91x10¹⁷ Hz
From Mosley's law,
√𝜈 =a(Z -b)
→9.44x10⁸ =5x10⁷(Z -1.356)
{The values of constants a and b are taken from the previous problem as directed}
→Z -1.356 =18.87
→Z =18.87 +1.356 
       ≈20
So this is the atomic number of the element calcium.             




 

       16. The Kᵦ X-rays from certain elements are given below. Draw the Mosley-type plot of √𝜈 versus Z for Kᵦ radiation.
Element             Na        P        Ca      Mn      Zn        Br
Energy (keV)   0.858    2.14    4.02    6.51    9.57    13.3           

ANSWER: Since the energy of a photon is given as 
E =h𝜈
→𝜈 =E/h
→√𝜈 =√(E/h)
From the given data, we have the following data for plotting:--
Element             Ne        P        Ca      Mn      Zn        Br
Energy (keV)   0.858    2.14    4.02    6.51    9.57    13.3 
√𝜈 (x10⁸)           4.55    7.19     9.85   12.54  15.20  17.92
Z                         10       15        20       25       30       35 

 We plot the graph as follows:--
Plot for Q-16

Z is plotted along the X-axis and √𝜈 along the Y-axis. As we see this graph is a straight line. 






       17. Use Mosley's law with b =1 to find the frequency of the Kₐ X-ray of La (Z =57) if the frequency of Kₐ X-ray of Cu (Z =29) is known to be 1.88x10¹⁸ Hz.             

ANSWER: Given, b =1 and the frequency of Kₐ X-ray 𝜈 =1.88x10¹⁸ Hz for Cu (Z =29).
 From Mosley's law, 
√𝜈 =a(Z -b)
→√(1.88x10¹⁸) =a(29 -1)
→a =1.37x10⁹/28
      =4.9x10⁷
Hence the frequency of Kₐ X-ray of La (Z =57) will be given as
√𝜈 =a(Z -b)
→√𝜈 =4.9x10⁷(57 -1)
→√𝜈 =2.74x10⁹
→𝜈 =7.50x10¹⁸ Hz.
           




 

       18. Kₐ and Kᵦ X-rays of molybdenum have wavelengths 0.71 Å and 0.63 Å respectively. Find the wavelength of Lₐ X-ray of molybdenum.          

ANSWER: From the given data, the wavelengths of Kₐ and Kᵦ X-rays of molybdenum are 0.071 nm and 0.063 nm respectively. Since the Kₐ X-ray results from the transition of an electron from the L shell to the K shell, the energy difference between the K shell and the L shell here is
E(L) -E(K) =hc/λ
             =1242/0.071 eV
             =17.5 keV
The Kᵦ X-ray results from the transition of an electron from the M shell to the K shell, hence the energy difference between the K shell and M shell here is
E(M) -E(K) =1242/0.063 eV
                  =19.7 keV
Hence the energy difference between the L shell and the M shell is
E(M) -E(L) =19.7 -17.5 =2.2 keV

We know that the Lₐ X-ray is emitted due to the transition of an electron from the M shell to the L shell. Hence the energy of a photon of the Lₐ X-ray here is
E =E(M) -E(L) =2.2 keV
Hence its wavelength is
𝜆 =1242/(2.2x10³) nm
   =0.564 nm
   =5.64 Å.
          




 

       19. The wavelengths of Kₐ and Lₐ X-rays of a material are 21.3 pm and 141 pm respectively. Find the wavelength of Kᵦ X-ray of the material.            

ANSWER: The wavelength λ of Kₐ X-ray =21.3 pm =0.0213 nm
Hence E(L) -E(K) =1242/0.0213 eV 
                             =58.3 keV
The wavelength λ of Lₐ X-ray =141 pm =0.141 nm.
Hence E(M) -E(L) =1242/0.141 eV
                             =8.81 keV

The energy of Kᵦ X-ray =E(M) -E(K)
              ={E(M) -E(L)} +{E(L) -E(K)}
              =8.81 +58.3 =67.11 keV.

So the wavelength of Kᵦ X-ray
  λ =1242/(67.11x10³) nm
     =0.0185 nm
     =18.5 pm.
         




 

       20. The energy of a silver atom with a vacancy in the K shell is 25.31 keV, the L shell is 3.56 keV, and the M shell is 0.530 keV higher than the energy of the atom with no vacancy. Find the frequency of Kₐ, Kᵦ, and Lₐ X-rays of silver.             

ANSWER: From the given data it implies that 25.31 keV of energy is required to separate widely apart an electron from the K shell of the silver atom, 3.56 keV from the L shell, and 0.530 keV from the M shell. Thus the energy difference between the K shell and the L shell is 
E(L) -E(K) =25.31 -3.56 =21.75 keV, 
Similarly, 
E(M)  -E(L) = 3.56 -0.53 =3.03 keV, and
E(M) -E(K) =25.31 -0.53 =24.78 keV. 

Hence the frequency of Kₐ X-ray
     ={E(L) -E(K)}/h
     =21.75x10³/4.14x10⁻¹⁵ Hz
     =5.25x10¹⁸ Hz.

The frequency of Kᵦ X-ray
         ={E(M) -E(K)}/h
         =24.78x10³/4.14x10⁻¹⁵ Hz
         =5.98x10¹⁸ Hz.

The frequency of Lₐ X-ray of silver
             ={E(M) -E(L)}/h
             =3.03x10³/4.14x10⁻¹⁵ Hz
             =7.32x10¹⁷ Hz.  

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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


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CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics




CHAPTER- 13 - Fluid Mechanics


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