Question

A tritium gas target is bombarded with a beam of monoenergetic protons of kinetic energy $K_1=3MeV$. The KE of the neutron emitted at ${30}^o$ to the incident beam is $K_2$? Find the value of $\frac{K_1}{K_2}=\frac{x}{2}$ (approximately in whole number). Find $x$.

Atomic masses are $H^1=1.007276amu,n^1=1.008665amu{,}_1{H}^3=3.016050amu{,}_{2}H{e}^3=3.016030amu$.

Answer 1

Here the equation of nuclear reaction will be, ${}_1{p}^1{+}_1{H}^3{\to }_{1}H{e}^3{+}_0{n}^1+Q$

where $Q=(m_p+m_H)-(m_{He}+m_n)=(1.007276+3.016050)-(3.016030+1.008665)$

$=-0.00137amu=-0.00137\times 931=-1.274MeV$

The kinetic energy of emitted neutron is: $K_2=\sqrt{u\pm (u^2+v{)}^{1/2}}$

where $u=\frac{\sqrt{m_p{m}_n{K}_1}}{m_{He}+m_n}\cos \theta =\frac{\sqrt{1.007276\times 1.008665\times 3}}{3.016030+1.008665}\cos 30=0.375$

and $v=\frac{Q{m}_{He}+K_1(m_{He}-m_P)}{m_{He}+m_n}=\frac{(-1.274)3.016030+3(3.016030-1.007276)}{3.016030+1.008665}=0.542$

Now, $K_2=\sqrt{0.375\pm ({0.375}^2+0.542{)}^{1/2}}=1.2MeV$ (- value is invalid because KE could not imaginary number )

So, $\frac{K_1}{K_2}=\frac{3}{1.2}=2.5$