Question

The nucleus of the polonium isotope 214 Po (mass 214 u) is radioactive and decays by emitting an alpha particle (a helium nucleus with mass 4 u). Laboratory experiments measure the speed of the alpha particle to be 1.97×107 m/s.

Assuming the polonium nucleus was initially at rest, what is the recoil speed of the nucleus that remains after the decay?

Answer 1

368224.29906

Step-by-step explanation:

Answer 2

368224.29906 m/s

Step-by-step explanation:

M = Mass of Polonium nucleus = 214 u

V = Velocity of nucleus

m = Mass of Helium nucleus = 4 u

v = Velocity of alpha particle =
`1.97* 10^7\ m/s`

In this system the momentum is conserved

`MV+mv=0\\\Rightarrow MV=-mv\\\Rightarrow 214V=-4* 1.97* 10^7\\\Rightarrow V=(-4* 1.97* 10^7)/(214)\\\Rightarrow V=368224.29906\ m/s`

The recoil speed of the nucleus that remains after the decay is 368224.29906 m/s