Question

A nucleus of mass number A originally at rest, emits
alpha particle with speed v. The recoil speed of the
daughter nucleus is :-
40
A+ 4
V
(3) -4
A+ 4
40
(1)
A-4​

Answer 1

Explanation:

It is given that, a nucleus of mass number A originally at rest, emits  alpha particle with speed v. The emission can be understand as follows :

$^4X\rightarrow^{A-4}Y+\alpha ^4$

Where

$^4X$ is the parent nucleus

$^{A-4}Y+\alpha ^4$ are the daughter nucleus.

Using the conservation of linear momentum as :

$0=m_Yv_Y+m_\alpha v_\alpha$ (initially, the nucleus is at rest)

Mass of Y is (A-4) and the mass of $\alpha$ is 4. So,

$0=(A-4)v_Y+4 v$ (since, speed is alpha particle is v)

$v_Y=\dfrac{-4v}{A-4}$

Negative sign shows the recoil speed of the daughter nucleus. Hence, this is the required solution.