Question

The wavelength of scattered radiation when it undergoes compton scattering at an angle of ${60}^o$ by graphite is $2.54\times {10}^{-11}$m, then the wavelength of incident photon is

• A. $4.2\times {10}^{-11}m$
• B. $1.12\times {10}^{-11}m$
• C. $1.21\times {10}^{-11}m$
• D. $2.42\times {10}^{-11}m$

Answer 1

Answer: (D)

$2.42\times {10}^{-11}m$

Compton formula
$\Delta \lambda ={\lambda }_f-{\lambda }_i=\frac{h}{m_{e}c}(1-cos\theta )$

$2.54\times {10}^{-11}-{\lambda }_1=\frac{6.63\times {10}^{-34}}{9.1\times {10}^{-31}\times 3\times {10}^8}(1-cos{60}^{\circ })$

${\lambda }_i=2.54\times {10}^{-11}-\frac{6.63\times {10}^{-34}}{9.1\times {10}^{-31}\times 3\times {10}^8}(1-1/2)(\because cos{60}^{\circ }=1/2)$

$=2.54\times {10}^{-11}-0.12\times {10}^{-11}$
$=2.42\times {10}^{-11}m$
So, the answer is option (D).