KCET · Physics · Dual Nature of Matter
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
- A 1
- B \(1 / 4\)
- C \(1 / 2\)
- D \(3 / 4\)
Answer & Solution
Correct Answer
(D) \(3 / 4\)
Step-by-step Solution
Detailed explanation
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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