Question

Q. DERIVATION OF EQUATION:-
$derive \: the \: equation \: \\ e = {mc}^{2}$
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Answer 1

Derivation:

$\qquad : \implies \sf M_m = M_n \bigg(1 + \dfrac{v^2}{2c^2} \bigg)$

Where,

• $\sf M_m$ = Total mass.
• $\sf M_r$ = Rest mass.

$\qquad : \implies \sf M_m = M_r + m_2 \dfrac{v^2}{2c^2}$

Subtracting $\sf M_r$ from both sides.

Let relative mass be m.

$\quad \sf Hence, \: M_m - M_r = m$

$\qquad : \implies \sf m = m_r \dfrac{v^2}{2c^2}$

$\qquad : \implies \sf \dfrac{\dfrac{1}{2} mrv^2}{c^2}$

We know that,

$\quad \sf Kinetic \: energy \: (KE) = \dfrac{1}{2} mv^2$

Where,

• m = $\sf m_r$ in this case.

Therefore,

$\qquad : \implies \sf m = \dfrac{e}{c^2}$

$\qquad \large \dag {\underline{\boxed{\sf e = mc^2}}}$

Hence derived. ✔︎