Question

A radioactive atom of mass 226 amu
shoots out an alpha particle (mass = 4
amu) with a speed of 1.5 x 107 ms. The
speed of the residual atom is​

Answer 1

$\huge{\bf{\blue{\fbox{\underline{\color{red}{Answer}}}}}}$

Conservation of total momentum of system implies:

P

1

(Momentum of alpha particle) + P

2

(Momentum of daughter nucleus) =0 (Initial momentum)

Thus,

$Thus, P1=−P2$

M= mass of daughter nucleus = 214 amu

m= mass of alpha particle = 4 amu

E is the energy of daughter nucleus in MeV

$Now , 6.7 \: MeV=2mP12 \: , for \: \: alpha \: \: particle$

$E=2MP22, for \: \:  daughter \: \: nucleus$

Dividing the above two equations, we get

$6.7E \: \: =Mm \: \:  (as P1=−P2 or P12=P22)</p><p></p><p>Therefore, \: \:  E= \: \: 6.7×2144=0.125</p><p></p><p>$