What is the recoil velocity of the nucleus that remains after the decay?

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.

When answering the question Assuming the polonium nucleus was initially at rest,

Answer 1

Conservation of momentum.  Mass x velocity of ejected particle must be equal to mass x velocity of the recoiling nucleus.

That's all!

Answer 2

it is important to note explicitly that the speed of the alpha particle is about 6% of the speed of light, so that ignoring relativistic effects are probably ok for this problem

then, use conservation of momentum, recognizing that there is zero momentum before the decay, and that there will be zero momentum after the decay

so, you know that:

m(alpha)v(alpha)=-m(remaining nucleus)v(remaining nucleus)

or that

v(rem nuc)= - m(alpha)/m(rem nuc) v(alpha)

the mass ratio you need is 4/210 (remember to subtract the mass of the alpha particle)...the rest is plug and chug

Answer 3

Let's assumes that the nucleus is free.

The momentum of both, the nucleus and the alpha particle are the same, because both were at rest in the beginning. The remaining nucleus has the mass 210u (it's lead Pb-210, by the way) that is about 52.5 times the mass of the alpha particle, thus the alpha particle has 52.5 times the velocity of the nucleus. The velocity of the nucleus is about 375 000 m/s.