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:

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\frac{v}{4}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{4 v}{238}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

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

Suggest Corrections

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:

Q. At

^{238} U

nucleus decays by emitting an alpha particle of speed

v

m/s. The recoil speed of the residual nucleus is (in m/s)

Q. When

U^{238}

nucleus originally at rest, decays by emitting an alpha particle having a speed

u

, the recoil speed of the residual nucleus is

Q. When a

U^{238}

nucleus, originally at rest, decays by emitting an

α

-particle, say with speed of v m/sec, the recoil speed of the residual nucleus is :(in m/sec.)

Q. A

U - 238

nucleus originally at rest, decays by emitting an

α -

particle, say with a velocity of

v m / s

. The recoil velocity in

(m s^{- 1})

of the residual nucleus is

Join BYJU'S Learning Program

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

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