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A proton moving with a momentum $p_{1}$ has a kinetic energy $1 / 8$ 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
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The correct answer is:
$3 / 4$
For proton, $v^{2}=\frac{c^{2}}{4}$ $\left(\because v=\frac{c}{2}\right)$
$\begin{aligned}
p_{1} &=\sqrt{2 m E_{k}} \\
&=\sqrt{2 \times m \times \frac{1}{8} m c^{2}} \quad\left(\because E_{k}=\frac{1}{8} m c^{2}\right) \\
p_{1} &=\frac{m \varepsilon}{2}
\end{aligned}$
For photon,
$\begin{aligned}
& E=\mathrm{KE} \\
& \frac{h c}{\lambda}=\frac{1}{8} m c^{2} \\
\Rightarrow \quad & p_{2}=\frac{h}{\lambda}=\frac{m c}{8} \\
& \therefore \quad \frac{p_{1}-p_{2}}{p_{1}}=\frac{\frac{m c}{2}-\frac{m c}{8}}{\frac{m c}{2}}=\frac{4-1}{8} \times 2=\frac{3}{4}
\end{aligned}$
$\begin{aligned}
p_{1} &=\sqrt{2 m E_{k}} \\
&=\sqrt{2 \times m \times \frac{1}{8} m c^{2}} \quad\left(\because E_{k}=\frac{1}{8} m c^{2}\right) \\
p_{1} &=\frac{m \varepsilon}{2}
\end{aligned}$
For photon,
$\begin{aligned}
& E=\mathrm{KE} \\
& \frac{h c}{\lambda}=\frac{1}{8} m c^{2} \\
\Rightarrow \quad & p_{2}=\frac{h}{\lambda}=\frac{m c}{8} \\
& \therefore \quad \frac{p_{1}-p_{2}}{p_{1}}=\frac{\frac{m c}{2}-\frac{m c}{8}}{\frac{m c}{2}}=\frac{4-1}{8} \times 2=\frac{3}{4}
\end{aligned}$
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