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 4.01×10-13 J. If after the decay, the alpha particle is observed to move in the positive x direction. After the decay, what direction did the thorium nucleus move? Complete the following statement with less than, greater than or equal to. After the decay, the kinetic energy of the thorium nucleus was _________ the kinetic energy of the alpha particle.

Answer 1

Solution,

Using law of connservation of momentum

(m1 V1) + (m2 V2) = 0

0.5 m1v1^2 = 4.01 x 10^-13

0.5 x 6.64 x 10^-27 x V1^2 = 4.01 x 10^-13

V1 = 1.1 x 10^7 m/s

(6.64 x 10^-27 x 1.10 x 10^7)+(3.89 x 10^-25 x V2) = 0

V2 = -1.87 x 10^5 m/s

Kinetic energy of thorium,

KE = 0.5*m2*V2^2

= 0.5 x 3.89*10^-25 x (1.87 x 10^5)^2 = 6.84 x 10^-15 J

Since, (4.01 x 10^-13) > (6.84 x 10^-15)

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

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