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)

- 4 v / 234

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v / 4

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- 4 v / 238

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4 v / 238

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Solution

The correct option is A $- 4 v / 234$
From the conservation of angular momentum,
$m_{1} v_{1} = m_{2} v_{2} + m_{3} v_{3} a s v_{1} = 0 m / s \Rightarrow m_{2} v_{2} = m_{3} v_{3} A T Q, v_{2} = v a n d m_{2} = 4 a n d m_{3} = 234. \Rightarrow 4 v = (234) (v_{3}) \Rightarrow v_{3} = \frac{(- 4 v)}{234}$
Therefore, the recital speed of the residual nucleus will be,
$v_{3} = \frac{(- 4 v)}{234}$

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At 238U nucleus decays by emitting an alpha particle of speed v m/s. The recoil speed of the residual nucleus is (in m/s)

The correct option is A −4v/234From the conservation of angular momentum,m1v1=m2v2+m3v3asv1=0m/s⇒m2v2=m3v3ATQ,v2=vandm2=4andm3=234.⇒4v=(234)(v3)⇒v3=(−4v)234Therefore, the recital speed of the residual nucleus will be, v3=(−4v)234

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)