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 = Ei = 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 = Ef = ?
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
1f′ = 1f + hm₀c² (1 - cosፀ)
As ፀ = 90° so cos90° = 0 So,
1f′ = 1f + hm₀c²
Now dividing both side of the equation by h we get
1hf′ = 1hf + 1m₀c² ------(1)
Here hf' = Ef = ? , hf = Ei = 50 keV (given) and the value of m0c2 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 = 81.99x10⁻¹⁵1.6x10⁻¹⁹ 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)
1Ef = 150keV + 1512keV
1Ef = 56225600keV
by flipping the fraction on both sides of the equation we get
Ef1 = 25600 keV562
or
Ef′ = 45.55 keV -------------------Ans.
************************************
************************************
Shortcut Links For
1. Website for School and College Level Physics 2. Website for School and College Level Mathematics 3. Website for Single National Curriculum Pakistan - All Subjects Notes
© 2022-Onwards by Academic Skills and Knowledge (ASK)
Note: Write me in the comment box below for any query and also Share this information with your class-fellows and friends.
1. Website for School and College Level Physics
2. Website for School and College Level Mathematics
3. Website for Single National Curriculum Pakistan - All Subjects Notes
© 2022-Onwards by Academic Skills and Knowledge (ASK)
Note: Write me in the comment box below for any query and also Share this information with your class-fellows and friends.
0 Comments
If you have any QUESTIONs or DOUBTS, Please! let me know in the comments box or by WhatsApp 03339719149