Question

In a nuclear reactor, a very slow-moving neutron is absorbed by a stationary boron atom. The equation for the nuclear reaction is
1n + 10B → 7Li + 4He
0 5 3 2
After the reaction, the speed of the helium atom is 9.1 x 106 ms-1 and the kinetic energy of the neutron is approximately zero.
Given:
mn = 1.008665 u;
mB = 10.0130 u;
mLi = 7.0160 u;
mHe = 4.0026 u.
Calculate
a) the kinetic energy of the lithium atom after the reaction.
b) the reaction energy in MeV and Joule.

Answer 1

(a) Since the neutron and boron are both initially at rest. The total momentum before the reaction is zero and afterward also it is zero. Therefore MLinLi=MHevHe

We solve this for vLi and substitute it into the equation for kinetic energy. We can use classical kinetic energy with little error, rather than relativistic formulas, because nHe=9.30×106ms−1 is not close to the speed of light c and nLi will be even less since nHe=9.30×106ms−1. Thus we can write

KLi=21MLivLi2=21MLi(MLiMHevHe)2=2MLiMHe2vHe2

We put in numbers, changing the mass in u to kg and recalling that 1.60×10−13J=1MeV

K

hope this helps you...