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maria [59]
3 years ago
15

At what speed, as a fraction of c , is a particle's total energy twice its rest energy

Physics
1 answer:
WINSTONCH [101]3 years ago
3 0
The rest energy of a particle is
E_0=m_0 c^2
where m_0 is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
E=mc^2 =  \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
E=2E_0
which means:
\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2
\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2
From which we find the ratio between the speed of the particle v and the speed of light c:
\frac{v}{c}=  \sqrt{1- (\frac{1}{2})^2 }  =0.87
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
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