Question

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

Answer 1

Let a particle of mass m moves with speed v .

then, total energy of particle is given by,

$E=\frac{mc^2}{\sqrt{1-\beta^2}}$

where, $\beta$ is ratio of speed of particle to velocity of light in vaccum.

and now rest - mass energy , R.E = mc²

but energy of particle is exactly twice of its rest mass energy.

so, E = 2R.E

or, $\frac{mc^2}{\sqrt{1-\beta^2}}=2mc^2$

or, $\frac{1}{\sqrt{1-\beta^2}}=2$

squaring both sides,

or, $\frac{1}{1-\beta^2}=4$

or, $\frac{1}{4}=1-\beta^2$

or, $\beta=\sqrt{\frac{3}{4}}$

now, speed of particle is v = $\beta c$

= √(3/4) × 3 × 10^8 m/s

= 2.598 × 10^8 m/s

hence, speed of particle is 2.598 × 10^8 m/s

Answer 2

Answer: 2.596 x 10⁸ ms⁻¹

mc²/√1-β² = Total Energy (T)

Rest mass energy is (R) = mc² (Einstein's law of restitution) where c = 299792458 ms⁻¹,

speed of light m = mass of particle, which is later rendered null.

Given -- T=2R ----- 1

Equating the value 1 , ∴ we get ,

mc²/√1-β² = 2 mc²

∴ 1/√1-β² = 2

Therefore, β = √3/4 = ratio of velocity of particle in vacuum to velocity of light in vacuum,

v/c = √3/4 ----where v is the velocity of particle,

v = √3/4 c = √3/4 x 299792488 = 259627884.491 ms⁻¹