Question

In a Compton effect experiment, the wavelength of incident photons is 3$A^0$.If the incident radiation is scattered through ${60}^0$ , the wavelength of scattered radiation is nearly (given$h=6.62\times {10}^{-34}Js$, $m_o=9.1\times 0^{-31}$ kg, $c=3\times {10}^8$ m/s)

• A. 3.024 $A^0$
• B. 3.012 $A^0$
• C. 3.048$A^0$
• D. 2.988 $A^0$

Answer 1

The correct option is B 3.012 $A^0$

${\lambda }^{\prime }-\lambda =\frac{h}{m_{e}c}(1-cos\theta )$

${\lambda }^{\prime }=\lambda +\frac{6.62\times {10}^{-34}}{9.1\times {10}^{-31}\times 3\times {10}^8}(1-cos{60}^{\bullet })$

$=\lambda +0.0121A^0$
$=3A^{\bullet }+0.0121$
$=3.012A^{\bullet }$

So, the answer is option (B).