A 50 keV X-ray is scattered through an angle of 90°. What is the energy of X-ray after Compton Scattering? (Answer: 45.5 keV)



Given:

Energy of the incident X-ray = `\E_i` =  50 keV

Scattering angle = ፀ = 90°

∴ Plank's Constant = h =  6.626 ｘ 10⁻³⁴ m² kg s⁻¹

∴ Rest mass of the electron = m₀ = 9.11 ｘ 10⁻³¹ kg

∴ Speed of light = 3 ｘ 10⁸ m s⁻¹


To Find:

The energy of the X-ray after Compton scattering  = `\E_f` =  ?



Solution:

Let f is the frequency of the incident X-ray and f' after Compton scattering then use Compton's equation in term of frequencies

`\frac {1}{f'}` =  `\frac {1}{f}`   +  `\frac {h}{m₀c²}` (1 - cosፀ) 

As  ፀ = 90° so cos90° = 0 So,

`\frac {1}{f'}` =  `\frac {1}{f}`   +  `\frac {h}{m₀c²}`

Now dividing both side of the equation by h we get

`\frac {1}{hf'}` =  `\frac {1}{hf}`   +  `\frac {1}{m₀c²}` ------(1)

Here hf' = `\E_f` = ? , hf = `\E_i` = 50 keV (given) and the value of `\m_0c^2` are as follow:

m₀c² = 9.11 ｘ 10⁻³¹ kg ｘ (3 ｘ 10⁸ m s⁻¹)²

m₀c² = 9.11 ｘ 10⁻³¹ kg ｘ 9 ｘ 10¹⁶ m² s⁻²

m₀c² = 81.99 ｘ 10⁻¹⁵ J

converting into eV we know that [ ∴ 1 eV = 1.6 ｘ 10⁻¹⁹ J ]

m₀c² = `\frac {81.99 ｘ 10⁻¹⁵ }{1.6 ｘ 10⁻¹⁹ }` eV

m₀c² = 51.24375 ｘ 10⁴ eV

m₀c² = 512.4375 ｘ 10³ eV

or

m₀c² = 512 keV------------------(2)

Now putting all the corresponding values in equation No. (1)


`\frac {1}{E_f}` =  `\frac {1}{50 keV}`   +  `\frac {1}{512 keV}`

`\frac{1}{E_f}` =  `\frac {562}{25600 keV}`

by flipping the fraction on both sides of the equation we get

`\frac {E_f}{1}` =  `\frac {25600  keV}{562}`

or

`\E_f'` =  45.55 keV -------------------Ans.





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