Question

The q value for the 3he(n,p) reaction is 0.76 mev. calculate the nuclide mass of 3he

Answer 1

Answer:

$0.7822 \mathrm{MeV}$

Explanation:

Nuclear reaction:

${ }^{3} \mathrm{H}+\mathrm{p} \rightarrow{ }^{3} \mathrm{He}+\mathrm{n}-0.7637 \mathrm{MeV}.......(i)\\ { }^{3} \mathrm{H}-{ }^{3} \mathrm{He}=\mathrm{n}-\mathrm{p}-0.7637.......(ii)$

Then, Mass difference

$m_{\mathrm{n}}-m_{\mathrm{H}}=m_{\mathrm{H}^{3}}-m_{\mathrm{He}^{3}}-m_{\mathrm{e}}+0.7637......(iii)$

Here the masses are atomic.

Add and subtract $m_{e}$in the right hand side so that

$m_{\mathrm{n}}-m_{\mathrm{H}}=m_{\mathrm{H}^{3}}-m_{\mathrm{He}^{3}}-m_{\mathrm{e}}+0.7637 \mathrm{MeV} .....(iv)$

The masses are now atomic.

Now consider the decay

${ }^{3} \mathrm{H} \rightarrow{ }^{3} \mathrm{He}+\beta^{-}+v+18.5 \mathrm{keV}$........(v)

On the atomic scale

$m_{\mathrm{H}^{3}}-m_{\mathrm{He}^{3}}=18.5 \mathrm{keV}=0.0185 \mathrm{MeV}$.........(vi)

$\text { use (vi) in (iv) to find }$

$m_{\mathrm{n}}-m_{\mathrm{H}}=(0.0185+0.7637) \mathrm{MeV}$

$=0.7822 \mathrm{MeV}$

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