A nucleus ( mass no. 238 ) initially at rest into another nucleus ( mass no 234 ) emitting an α particle (mass no-4) with the speed of1.17×107m/sec. Find the recoil speed of the remaining nucleus.
the nucleus was at rest initially.
so, it has momentum =0
as,man of α-particle has speed V1=1.17×107m/s
from conservation of momentum m1v1+m2v2=0
v2=−m1v1m2=−4×1.17×107234=−2×105m/s
∴ recoil velocity of nucleus = 2×105m/s.