Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

A proton is bombarded on a stationary lithium nucleus. As a result of collision two $\;\alpha\;$ particles are produced . If direction of motion of $\;\alpha \;$ particle with initial direction of motion makes an angle $\;cos^{-1}(\large\frac{1}{4})\;$ find kinetic energy of striking proton . Given binding energies per nucleon of $\;Li^{7}\;$ & $\;He^{4}\;$ are $5.60 MeV$ and $7.06 MeV$ respectively


Can you answer this question?

1 Answer

0 votes
Answer : (b) $\;17.28\;MeV$
Q value of reaction is
Therefore , $Q=(2 \times 4 \times 7.06 -7 \times 5.6)$
$=17.28 MeV$
Where $\;K_{p}\;$ & $\;K_{\alpha}\;$ are kinetic energies of photon and $\;\alpha \;$ particle respectively . By conservation of linear momentum
$\sqrt{2m_{p} K_{p}}=2 \sqrt{2 m_{\alpha}K_{\alpha}} cos \theta$
$K_{p}=16 K_{\alpha} cos^{2} \theta =K_{\alpha}$
$K_{p}=17.28 \;MeV$
answered Mar 11, 2014 by yamini.v
edited Mar 13, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App