Home/Class 11/Physics/

## QuestionPhysicsClass 11

The velocity of a particle at which the kinetic energy is eqyal to its rest mass energy is
(A) $${\left(\frac{{{3}{c}}}{{{2}}}\right)}$$
(B) $${3}\frac{{{c}}}{{\sqrt{{{2}}}}}$$
(C) $$\frac{{{\left({3}{c}\right)}^{{{1}/{2}}}}}{{{2}}}$$
(D) $$\frac{{{c}\sqrt{{{3}}}}}{{{2}}}$$

The relativistic K.E. of a particle of rest mass $${m}_{{{0}}}$$ is given by
$${K}={\left({m}-{m}_{{{0}}}\right)}{c}^{{{2}}},$$ where$${m}=\frac{{{m}_{{{0}}}}}{{\sqrt{{{1}-\upsilon^{{{2}}}/{c}^{{{2}}}}}}}$$
Here, $${m}$$ is mass of particle moving with velocity $$\upsilon$$.
As $${K}.{E}.=$$ rest mass energy
$$\therefore{\left({m}-{m}_{{{0}}}\right)}{c}^{{{2}}}={m}_{{{0}}}{c}^{{{2}}}$$ or $${m}{c}^{{{2}}}={2}{m}_{{{0}}}{c}^{{{2}}}$$
$$\frac{{{m}_{{{0}}}}}{{\sqrt{{{1}-\upsilon^{{{2}}}/{c}^{{{2}}}}}}}={2}{m}_{{{0}}}$$
$${1}-\upsilon^{{{2}}}/{c}^{{{2}}}=\frac{{{1}}}{{{4}}}$$
$$\upsilon^{{{2}}}/{c}^{{{2}}}={3}/{4},\upsilon=\frac{{\sqrt{{{3}}}}}{{{2}}}{c}$$