Q85 / 120 · VITEEE 2007 · Physics
The velocity of a particle at which the kinetic energy is equal to its rest energy is
- A$\left(\frac{3 \mathrm{c}}{2}\right)$
- B$3 \frac{\mathrm{c}}{\sqrt{2}}$
- C$\frac{(3 \mathrm{c})^{\frac{1}{2}}}{2}$
- Correct answerD$\frac{\mathrm{c} \sqrt{3}}{2}$
Solution
Here, Kinetic energy = Rest energy we know that
Kinetic energy (Relativistic) $=\left(\mathrm{m}-\mathrm{m}_{0}\right) \mathrm{c}^{2}$
and Rest energy $=m_{0} c^{2}$
where $m_{0}=$ rest mass, $\mathrm{c}=$ velocity of light
Also, mass (m) of a particle moving with velocity $v$ is given by
$\mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\left(\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}\right)}}$
$\therefore\left(\mathrm{m}-\mathrm{m}_{0}\right) \mathrm{c}^{2}=\mathrm{m}_{0} \mathrm{c}^{2}$ provides
$\left(\frac{\mathrm{m}_{0}}{\sqrt{1-\left(\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}\right)}}-\mathrm{m}_{0}\right) \mathrm{c}^{2}=\mathrm{m}_{0} \mathrm{c}^{2}$
or, $\frac{1}{\sqrt{1-\left(\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}\right)}}-1=1$
or, $\frac{1}{\sqrt{1-\left(\frac{v^{2}}{c^{2}}\right)}}=2 \quad$ or, $\quad 1-\frac{v^{2}}{c^{2}}=\frac{1}{4}$ or, $\frac{v^{2}}{c^{2}}=1-\frac{1}{4}=\frac{3}{4}$
$\begin{array}{l}
\therefore \mathrm{v}=\frac{\sqrt{3}}{2} \mathrm{c} \\
\Rightarrow 1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}=\frac{1}{4} \\
\Rightarrow \frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}=1-\frac{1}{4}=\frac{3}{4} \Rightarrow \mathrm{v}^{2}=\frac{3}{4} \mathrm{c}^{2} \\
\Rightarrow \mathrm{v}=\sqrt{\frac{3}{2}} \mathrm{c}
\end{array}$
Hence, the correct answer is option: 'D'.