Question

If total energy of a particle is exactly twice of its rest mass energy. Calculate its speed

Answer 1

E=2mc. {c=v}
then, v=E/2m

Answer 2

Answer:

The speed of the particle is $v=\dfrac{\sqrt{3}}{2}$

Explanation:

Given that,

If total energy of a particle is exactly twice of its rest mass energy.

$E =2E_{0}$....(I)

Total energy of the particle is defined as:

$E = \dfrac{m_{0}c^2}{\sqrt{1-v^2}}$

Here, $m_{0}$= rest mass of the particle

c = speed of light

$v$= speed of the particle

Rest energy of the particle is

$E_{0} = m_{0}c^2$

Therefore, from equation (I)

$\dfrac{m_{0}c^2}{\sqrt{1-v^2}}= 2\times m_{0}c^2$

$\dfrac{1}{\sqrt{1-v^2}}=2$

$\sqrt{1-v^2}=\dfrac{1}{2}$

$1-v^2=\dfrac{1}{4}$

$v^2=\dfrac{3}{4}$

$v=\dfrac{\sqrt{3}}{2}$

Hence, The speed of the particle is $v=\dfrac{\sqrt{3}}{2}$