Question

The nucleus of a certain atom has a mass of 3.8 x 10^-25 kg and is at rest. The nucleus is radioactive and suddenly eject from itself a particle of mass 6.6 x 10^-27 kg and speed 1.5 x 10⁷ m/s. Find the recoil speed of the nucleus left behind...

Answer 1

Here we shall use the law of conservation of momentum.

Initial momentum = Final momentum

as the nucleus is initially at rest

Therefore, initial momentum = 0

So, Final momentum = $MV+mv=0$

or

$MV=-mv$

so,

$V=-\left(mv/M\right)$

here,

$m=6.6\times {10}^{-27}kg$

$v=1.5\times {10}^{7}m/s$

$M=3.8\times {10}^{-25}kg$

so,

velocity of the larger nucleus will be

$V=-(6.6\times {10}^{-27}\times 1.5\times {10}^7)/(3.8\times {10}^{-25})$

thus,

$|V|=2.6\times {10}^{5}m/s$