A $^{238} U$ nucleus decays by emitting an alpha particle of speed $^{'} v^{'} m s^{- 1}$ . The recoil speed of the residual nucleus (in $m s^{- 1}$ ) is:

$\frac{- 4 v}{234}$

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$\frac{v}{4}$

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$\frac{- 4 v}{238}$

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$\frac{4 v}{238}$

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Solution

The correct option is A
$\frac{- 4 v}{234}$

Initially $^{238} U$ nucleus was at rest and after decay its part moves in opposite direction.

According to conservation of momentum

$4 v + 234 V = 238 \times 0 \Rightarrow V = - \frac{4 v}{234}$

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Similar questions

A $^{238} U$ nucleus decays by emitting an alpha particle of speed $^{'} v^{'} m s^{- 1}$ . The recoil speed of the residual nucleus (in $m s^{- 1}$ ) is:

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A 238U nucleus decays by emitting an alpha particle of speed ′v′ ms−1. The recoil speed of the residual nucleus (in ms−1 ) is:

The correct option is A −4v234 Initially 238U nucleus was at rest and after decay its part moves in opposite direction. According to conservation of momentum 4v+234 V=238×0⇒V=−4v234

A $^{238} U$ nucleus decays by emitting an alpha particle of speed $^{'} v^{'} m s^{- 1}$ . The recoil speed of the residual nucleus (in $m s^{- 1}$ ) is:

The correct option is A
$\frac{- 4 v}{234}$

Initially $^{238} U$ nucleus was at rest and after decay its part moves in opposite direction.

According to conservation of momentum

$4 v + 234 V = 238 \times 0 \Rightarrow V = - \frac{4 v}{234}$