The rest energy of an electron is 0.511 , MeV | vedclass

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(C) The rest energy of the electron is given by $E_0 = m_0 c^2 = 0.511 \, MeV$.When the electron is accelerated to a velocity $v = 0.5 \, c$,its total

Vedclass · Accepted Answer

$0.079$

Vedclass · Answer

(C) The rest energy of the electron is given by $E_0 = m_0 c^2 = 0.511 \, MeV$.When the electron is accelerated to a velocity $v = 0.5 \, c$,its total energy $E$ is given by the relativistic formula:$E = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}}$Substituting $v = 0.5 \, c$:$E = \frac{0.511}{\sqrt{1 - (0.5)^2}} = \frac{0.511}{\sqrt{1 - 0.25}} = \frac{0.511}{\sqrt{0.75}}$$E = \frac{0.511}{0.866} \approx 0.590 \, MeV$The change in energy $\Delta E$ is the difference between the total energy and the rest energy:$\Delta E = E - E_0 = 0.590 \, MeV - 0.511 \, MeV = 0.079 \, MeV$.

The rest energy of an electron is $0.511 \, MeV$. The electron is accelerated from rest to a velocity $0.5 \, c$. The change in its energy will be ............ $MeV$.

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