$(a)\;2.04 MeV\qquad(b)\;1.02 MeV\qquad(c)\;0.04 MeV\qquad(d)\;3.2 MeV$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer : (b) $\;1.02 MeV$

Explanation :

Since neutron and boron are both initially at rest , total momentum before reaction is zero , and after ward is also zero

$H_{Li} \nu_{Li}=M_{He} V_{Li}$

We solve this for $\;V_{Li}\;$ and substitute it in to equation for kinetic energy .We can use classical kinetic energy .

Thus $\;K_{Li}=\large\frac{1}{2}\;M_{Li}V_{Li}^{2}=\large\frac{1}{2} M_{Li}\;(\large\frac{M_{He} V_{He}}{M_{Li}})^2$

$=\large\frac{M_{He}^2 V_{He}^2}{2 M_{Li}}$

$K_{Li}=\large\frac{(4.0026 u)^2\;(1.66\times 10^{-27}\;Kg/w)\;(9.30 \times10^{6} m/s)^2}{2(7.016 u)\;(1.66\times10^{-27} Kg/u)}$

$=1.64 \times 10^{-13} J=1.02 MeV\;.$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...