Question

A u-238 nucleus originally at rest,decays by emitting an alpha particle say with a velocity of v m/s.the recoil velocity of the residual nucleus is

Answer 1

Hey mate ^_^

Here is ur answer... ⏬⏬

Recoil of the residual nucleus is -4v/234..

This is by momentum conservation..

Final momentum=initial momentum

4v+234v'=0

4v=-234v'

v'=-4v/234 m/s


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Answer 2

Answer:

The recoil velocity of the residual nucleus is equal to -4v/234ms⁻¹.

Explanation:

Let's assume that,  m is the mass of proton

Mass of  alpha particle $m_a}=4m$

Mass of residual nucleus $m_{r}=(238-4)m=234m$

From law of momentum conservation

               $m_{a}v_{a}+ m_{r}v_{r}=0$                        ...........(1)

Fill the value of $v_{a}=v$ and value of mass in eq. (1),

Hence        $4m(v)+234m(v_{r})=0\\\\ 234m(v_{r})= -4mv\\\\ v_{r}=\frac{-4v}{234}$

where $v_{r}$  is the recoil velocity of the nucleus.