Ask Questions, Get Answers

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

A nucleus with mass number 200 initially at rest emits and $\;\alpha-\;$ particle . If Q value of reaction is 4.2MeV .Then kinetic energy of $\;\alpha-\;$ particle will be

$(a)\;4.2 MeV\qquad(b)\;4.116 MeV\qquad(c)\;3.96 MeV\qquad(d)\;1.9 MeV$

Can you answer this question?

1 Answer

0 votes
Answer : (b) $\;4.166 MeV$
Explanation :
Given that $\;K_{1}+K_{2}=4.2 MeV----(1)$
From conservation of linear momentum
or $\; \sqrt{2 K_{1} (196 m)}=\sqrt{2 K_{2} (4m)} \qquad \; [P=\sqrt{2km}]$
Solving (1) & (2)
We get
$K_{2}=KE \;of\;\alpha-particle=4.116MeV\;.$
answered Feb 28, 2014 by yamini.v

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