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...
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=−(mv/M)
here,
m=6.6×10−27kg
v=1.5×107m/s
M=3.8×10−25kg
so,
velocity of the larger nucleus will be
V=−(6.6×10−27×1.5×107)/(3.8×10−25)
thus,
|V|=2.6×105m/s