The rest mass energy of an electron is 0.51 | vedclass

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(B) Given: Rest mass energy $E_0 = m_0c^2 = 0.51 \ MeV$ and velocity $v = 0.8 \ c$.The kinetic energy $(K.E.)$ of a relativistic particle is given by

Vedclass · Accepted Answer

$0.34$

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(B) Given: Rest mass energy $E_0 = m_0c^2 = 0.51 \ MeV$ and velocity $v = 0.8 \ c$.The kinetic energy $(K.E.)$ of a relativistic particle is given by $K.E. = mc^2 - m_0c^2$.The relativistic mass $m$ is given by $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$.Substituting $v = 0.8 \ c$:$m = \frac{m_0}{\sqrt{1 - (0.8)^2}} = \frac{m_0}{\sqrt{1 - 0.64}} = \frac{m_0}{\sqrt{0.36}} = \frac{m_0}{0.6}$.Now,the total energy $E = mc^2 = \frac{m_0c^2}{0.6} = \frac{0.51 \ MeV}{0.6} = 0.85 \ MeV$.Therefore,$K.E. = E - E_0 = 0.85 \ MeV - 0.51 \ MeV = 0.34 \ MeV$.

The rest mass energy of an electron is $0.51 \ MeV$. If this electron is moving with a velocity of $0.8 \ c$ (where $c$ is the velocity of light in vacuum),then the kinetic energy of the electron is ........... $MeV$.

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