Question

The velocity of a particle at which the kinetic energy is equal to its rest energy is

• A. $\bigg( \frac{3c}{2}\bigg)$
• B. $3 \frac{c}{\sqrt{2}}$
• C. $\frac{(3c)^{\frac{1}{2}}}{2}$
• D. $\frac{c\sqrt{3}}{2}$

Answer 1

Answer: (D)

$\frac{c\sqrt{3}}{2}$

Answer 2

The relativistic kinetic energy of a particle of rest mass $m_{0}$ is given by $\quad K=\left(m-m_{0}\right) c^{2}$
$m=\frac{m_{0}}{\sqrt{1-\left(v^{2} / c^{2}\right)}}$, where $m$ is the mass of
the particle moving with velocity $v$.
$\therefore K=\left[\frac{m_{0}}{\sqrt{1-\left(v^{2} / c^{2}\right)}}-m_{0}\right] c^{2} z$
According to problem,
kinetic energy $=$ rest energy
$\therefore\left[\frac{m_{0}}{\sqrt{1-\left(v^{2} / c^{2}\right)}}-m_{0}\right] c^{2}=m_{0} c^{2}$
or $\frac{m_{0} c^{2}}{\sqrt{1-\left(v^{2} / c^{2}\right)}}=2 m_{0} c^{2}$
or $\frac{1}{1-\left(v^{2} / c^{2}\right)}=4$
or $4 v^{2} / c^{2}=3$
$\therefore v=\frac{\sqrt{3} c}{2}$