Question

When a $U^{238}$ nucleus originally at rest, decay by emitting an alpha particle having a speed $1.5\times {10}^7{m}^{-1}$. Then calculate the kinetic energy of the residual nucleus.

Answer 1

Mass of $\alpha$ particle, $m_{\alpha }=4u$

Mass of nucleus after fission, $m_n=234u$

From conservation of momentum

Final momentum = initial momentum

$m_{\alpha }{v}_{\alpha }+m_n{v}_n=0$

$\Rightarrow 4\times 1.5\times {10}^7+234v_n=0$

$\Rightarrow v_n=2.56\times {10}^{5}m{s}^{-1}$

Hence, velocity of nucleus is $2.56\times {10}^{5}m{s}^{-1}$