Question

A proton moving with a momentum $P_1$ has a kinetic energy $\frac{1}{8}$ th of its rest mass energy. Another light photon having energy equal to the kinetic energy of the possesses a momentum $P_2.$ Then the ratio $\frac{P_1-P_2}{P_1}$ is equal to

• A. 1
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

Answer: (D)

For proton $v^2=\frac{c^2}{4}v=\frac{c}{2}$
$P=\sqrt{2m{E}_k}$
$=\sqrt{/2m\frac{1}{/8_4}m{c}^2}$
$P_1=\frac{mc}{2}$
For photon
$E_{\text{photon }}=K{E}_{\text{proton }}$
$\frac{h/c}{\lambda }=\frac{1}{8}m{c}^{/2}$
$P_{\text{photon }}=\frac{h}{\lambda }=\frac{mc}{8}=P_2$
$\frac{P_1-P_2}{P_1}=\frac{\frac{/mc}{2}-\frac{/mc}{8}}{\frac{/mc}{2}}$
$=\frac{\frac{1}{2}-\frac{1}{8}}{\frac{1}{2}}=\frac{\frac{4-1}{8}}{\frac{1}{2}}$
$=\frac{3\times 2}{8}=\frac{3}{4}$