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  • A tritium gas target is bombarded with a beam of mono-energetic protons of <img alt="\mathrm{K.E. 3 \: MeV}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BK.E.%203%20%5C%3A%20MeV

A tritium gas target is bombarded with a beam of mono-energetic protons of \mathrm{K.E. 3 \: MeV}. What is the \mathrm{K.E. } of the neutrons which are emitted at an angle \mathrm{30^{\circ} } with the incident beam?

Atomic masses are:\mathrm{ \begin{aligned} & H^1=1.007276 \mathrm{amu} \quad n^1=1.008665 \mathrm{amu} \\ \end{aligned} }

                             \mathrm{ \begin{aligned} & H_1^3=3.06056 \mathrm{amu} \text { and } H_e^3=3.016030 \mathrm{amu} . \end{aligned} }

 

Option: 1

1.20 \mathrm{MeV}


 


Option: 2

1.44 \mathrm{MeV}
 


Option: 3

1.60 \mathrm{MeV}
 


Option: 4

1.10 \mathrm{MeV}


Answers (1)

best_answer

The nuclear reaction in question can be written as-

\mathrm{P}_1^1+\mathrm{H}_1^3 \rightarrow \mathrm{He}_2^3+\mathrm{n}_0^1+\mathrm{Q}

when \mathrm{Q}is the energy released during the reaction given by

\mathrm{Q =\left[\left(m_p+m_H\right)-\left(m_{H e}+m_n\right)\right] \mathrm{amu} }

\mathrm{=(1.0072765+3.06056)-(3.016030-1.008665) }

\mathrm{ =-0.001369 \mathrm{amu}=-1.2745 \mathrm{MeV} }

From conservation of energy, we get

\mathrm{ \mathrm{k}_{\mathrm{p}} =\mathrm{k}_{\mathrm{n}}+\mathrm{k}_{\mathrm{He}}+\mathrm{Q} }

\mathrm{ \Rightarrow \mathrm{k}_{\mathrm{n}} =\mathrm{k}_{\mathrm{H}}-\mathrm{k}_{\mathrm{He}}-\mathrm{Q}=3-\mathrm{k}_{\mathrm{He}}+1.2745=4.2745-\mathrm{k}_{\mathrm{He}}}.......(1)

From conservation of linear momentum along and perpendicular to the direction of proton beam, we get

\mathrm{ \mathrm{p}_{\mathrm{H}} =\mathrm{p}_{\mathrm{n}} \cos 30^{\circ}+\mathrm{p}_{\mathrm{He}} \cos \theta }

\mathrm{\Rightarrow \mathrm{p}_{\mathrm{H}} =\frac{\sqrt{3}}{2} \mathrm{p}_{\mathrm{n}}+\mathrm{p}_{\mathrm{He}} \cos \theta}            ........(2)

and

\mathrm{ 0=p_{\mathrm{n}} \sin 30^{\circ}-p_{\mathrm{He}} \sin \theta }

\mathrm{\Rightarrow p_{\mathrm{He}} \sin \theta=\frac{p_{\mathrm{n}}}{2}}                               .........(3)

Using the formula for kinetic energy \mathrm{k=\frac{p^2}{2 m}} in the above equations (1)-(3), and putting the values of masses from the given data, we get,

\mathrm{ \Rightarrow \mathrm{k}_{\mathrm{n}}=1.44 \mathrm{MeV} }

Hence option 2 is correct.
 

Posted by

Divya Prakash Singh

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