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svp [43]
2 years ago
8

The nucleus of the polonium isotope 214 Po (mass 214 u) is radioactive and decays by emitting an alpha particle (a helium nucleu

s 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?
Physics
2 answers:
amid [387]2 years ago
6 0

Answer:

368224.29906 m/s

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\times 10^7\ m/s

In this system the momentum is conserved

MV+mv=0\\\Rightarrow MV=-mv\\\Rightarrow 214V=-4\times 1.97\times 10^7\\\Rightarrow V=\dfrac{-4\times 1.97\times 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

Maksim231197 [3]2 years ago
4 0

Answer:

368224.29906

Explanation:

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