A $^{238}U$ nucleus decays by emitting an $\alpha$ particle of speed $v\,m{s^{ - 1}}$. The recoil velocity of the residual nucleus is (in $m{s^{ - 1}}$)
A $^{238}U$ nucleus decays by emitting an $\alpha$ particle of speed $v\,m{s^{ - 1}}$. The recoil velocity of the residual nucleus is (in $m{s^{ - 1}}$)
$ - 4v/234$
$v/4$
$ - 4v/238$
$4v/238$
A $^{238}U$ nucleus decays by emitting an $\alpha$ particle of speed $v\,m{s^{ - 1}}$. The recoil velocity of the residual nucleus is (in $m{s^{ - 1}}$)
Initially $238\,U$ nucleus was at rest and after decay its part moves in opposite direction.
According to conservation of momentum
$4v + 234V = 238 × 0 $
==> $V = - \frac{{4v}}{{234}}$
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