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How much energy must a bombarding proton possess to cause the reaction ? $\;^{7}H_{3}+^{1}H_{1} \to _{4} Be^{7}+_{0}n^{1}$

$(a)\;2.2 MeV\qquad(b)\;1.7 MeV\qquad(c)\;1.9 MeV\qquad(d)\;1.6 MeV$

1 Answer

Answer : (c) $\;1.9 MeV$
Explanation :
Since mass of an atom include the masses of atomic electrons , the approprite no.of electron masses must be subtracted from given values
Reactans$\qquad \; \qquad \; \qquad \; \qquad \; \qquad \; \qquad \;$Products
$_{3}^{7}Li\;7.016 -3m_{e} \;_{4}^{7}Be \qquad \; \qquad \qquad 1.01693-4m_{e}$
$_{1}^{1}H\;1.078-1m_{e}\;_{0}^{1}n \qquad \qquad \qquad \qquad1.0866$
Total $\;8.023-4m_{e} \qquad \; Total \; 8.025-4m_{e}$
Mass defect = $\; 8.023-8.02559=-0.00176$
The Q value of reaction
$Q=-0.00179 u=-16.5 MeV$
The energy is supplied as KE of bombarding proton. The incident proton must have more than this energy because the system must possess some KE even after the reaction So , that momentum is conserved .
With momentum conservation taken into account the minimum KE that incident particle must have can be found with formula.
$=-1.89 MeV\;.$
answered Feb 23, 2014 by yamini.v

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