Ask Questions, Get Answers

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

A tritium gas target is bombareled with a beam of mono-energetic protons of K.E 3MeV . What is K.E of neutrons emitted at an angle $\;30^{0}\;$ with incident beam ? \[\] Atomic masses are $\;H^{1}=1.007276 amu$ \[\] $\;m^{1}=1.008665 amu$\[\] $\;H_{1}^{3}=3.06056 amu$\[\] and $\;He^{2}=3.016030 amu$

$(a)\;1.44 MeV\qquad(b)\;2.36 MeV\qquad(c)\;1.92 MeV\qquad(d)\;2.6 MeV$

Can you answer this question?

1 Answer

0 votes
Answer : (a) $\;1.44 MeV$
Explanation :
The nuclear reaction in question can be written as $\;P_{1}^{1}+H_{1}^{3} \to He_{2}^{3}+n_{0}^{1}+Q$
When Q is energy released during reaction given by
$=-0.001369 amu=-1.2745 MeV$
The kinetic energy of associated emitted motion is given by
$K_{n}=[u_{n}\pm (U_{n}^{2}+v^2)]---(1)$
where $\;u_{n}=\large\frac{(m_{p} m_{n} K_{p})^{\large\frac{1}{2}}}{m_{He}+m_{n}} cos \theta$
$u_{n} \approx =0.3753$
and $\;V=\large\frac{m_{He}Q+K_{p}(m_{He}-m_{p})}{m_{He}+m_{n}}$
$V \approx 0.5424 $
$K_{n}=[0.3753\pm 0.8266]$
$K_{n}=1.44 MeV\;.$
answered Feb 26, 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