Question

A uranium-238 nucleus at rest undergoes radioactive decay, splitting into an alpha particle (helium nucleus) with mass 6.64×10^-27 kg and a thorium nucleus with mass 3.89×10^-25 kg. The measured kinetic energy of the alpha particle is 7.11×10^-13 J.

1) After the decay, the kinetic energy of the thorium nucleus was _________ the kinetic energy of the alpha particle. (less than, greater than or equal to.)

Answer 1

Using law of connservation of linear momentum

Linear momentum before decay = Linear momentum after decay

0 = (m1*V1)+(m2*V2)

V1 is the speed of alpha particle

0.5*m1*v1^2 = 7.11*10^-13

0.5*6.64*10^-27*V1^2 = 7.11*10^-13

V1 = 1.46*10^7 m/s

m2 = 3.89*10^-25 kg

V2 = ?

then

0 = (6.64*10^-27*1.46*10^7)+(3.89*10^-25*V2)

V2 = -2.49*10^5 m/s

Kinetic energy of thorium is K2 = 0.5*m2*V2^2 = 0.5*3.89*10^-25*(2.49*10^5)^2 = 1.21*10^-14 J

Noe comparing

7.11*10^-13 > 1.21*10^-14 J

So After the decay, the kinetic energy of the thorium nucleus was less than the kinetic energy of the alpha particle.