A radioactive nucleus is at rest It disintegrates into three

A radioactive nucleus is at rest. It disintegrates into three particles. Two of them are detected, with masses and velocities shown below. What is the momentum of the third particle, which is known to have a mass of 12 times 10^-27 kg? How much energy was involved in the disintegration process?

Solution

Aplying law of conservation of momentum:

Momentum before disintegration = momentum after disintegration

( M1 + M2 + M3 ) ( 0 ) = M1 * V1 + M2 * V2 + M3 * V3

In the X direction:

0= ( 6 * 10 ^-27 * 6.6 * 10 ^ 6 ) + Momentum of third particle

Momentum of thrid particle in X - direction = - 39.6 * 10^-21

Velocity of the third particle in X direction = momentum in X direction / Mass = -39.6 * 10 ^-21 / ( 12 X 10^-27 ) = -3.3 * 10 ^6 m /s

In Y direction:

0 = ( 4.5 * 10^-27 * 8 * 10 ^ 6 ) + Momentum of the third particle

Momentum of third particle in Y direction= -36 X 10 ^ -21

Velocity of the third partcile in Y direction = Momentum in Y direction / Mass = 3 * 10^6 m /s

Therefore velocity of the third particle: Sqrt ( ( velocity in x- direction )^2 + ( velocity in Y - direction ) ^2 )) = 4.46 * 10 ^6 m/s

b. Energy involved in the disintegration process: = sum of energies of each of the particles

= I/2 * M1 * V1 * V1 + 1/2 M2 * V2 * V2 + 1/2 * M3 * V3 * V3

= 144 * 10 ^-15 + 130.68 * 10 ^-15 + 119.35 * 10^-15 = 394.03 * 10 -15 Joules

Energy involved = 394 X 10 ^ -15 Joules