In: Physics
Beryllium-8 is an unstable isotope and decays into two α particles, which are helium nuclei with mass 6.68×10−27kg. This decay process releases 1.5×10−14J of energy. For this problem, let's assume that the mass of the Beryllium-8 nucleus is just twice the mass of an α particle and that all the energy released in the decay becomes kinetic energy of the α particles.
If a Beryllium-8 nucleus is at rest when it decays, what is the speed of the α particles after they are released? (My answer = 1.5 x 10^6)
If the Beryllium-8 nucleus is moving in the positive x-direction with a speed of 1.0×106 m/s when it decays, what is the speed of the slower-moving α particle after it is released? Assume that the α particles move entirely in the x-direction.
If the Beryllium-8 nucleus is moving in the positive x-direction with a speed of 1.0×106 m/s when it decays, what is the speed of the faster-moving α particle after it is released? Assume that the α particles move entirely in the x-direction.
Here, Mass of alpha particle is
= 6.68*10-27 kg
The mass of Beryllium is
= 2*6.68*10-27 = 1.336*10-26 kg
The energy released = 1.5*10-14 J
a) In this case, the energy released is equally divided among the two daughter nuclei.
So, the energy got by each daughter nuclei is
0.75*10-14 J
Using E = mv2/2,
Ea = mav2/2
v2 = 2*Ea /ma = 2*0.75*10-14 /(6.68*10-27 ) = 2.2455*1012
v = 1.498*106 m/s
b) Here, the initial nucleus is moving forward with a velocity of 1.0*106 m/s
This will give some extra energy for the forward moving particle and some less energy to the backward moving particle.
The velocity of the particles add up. Relativistic effects are so small and can be neglected.
For the slower particle,
Here, the effective velocity is
V = V1 - V2 = 1*106 - 1.498*106 = - 0.498*106 m/s
The negative sign only represents that the direction of motion of the particle is in the -X axis.
c) For the faster moving particle,
Here, the effective velocity is
V = V1 + V2 = 1*106 + 1.498*106 = 2.498*106 m/s