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Chapter 11: Q. 68 (page 290)

The nucleus of the polonium isotope214Po(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 1.92脳107m/s. Assuming the polonium nucleus was initially at rest, what is the recoil speed of the nucleus that remains after the decay?

Short Answer

Expert verified

The recoil speed of the nucleus that remains after the decay is -3.59脳105m/s.

Step by step solution

01

Step 1. Given information is :Mass of214Po= 214 uMass of He = 4 uSpeed of alpha particle = 1.92×107 m/sPolonium nucleus initially at rest.

We need to find out the recoil speed of the nucleus that remains after the decay.

02

Step 2. Using Law of conservation of momentum to find out the recoil speed of the nucleus that remains after the decay. 

Initially, alpha particle and Nucleus both are at rest, so Pi = 0

And as per Law of conservation of momentum,

Pi=Pf0=mpvpi+m伪v伪ivpi=-m伪v伪impPuttingvalues,vpi=-(4u)(1.92脳107m/s)214uvpi=-3.59脳105m/s

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