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(A) The decay reaction is: $n ightarrow p + e^- + \bar{
u} + Q$.The mass defect $\Delta m$ is given by: $\Delta m = m_n - (m_p + m_e)$.Substituting

Vedclass · Accepted Answer

$0.51$

Vedclass · Answer

(A) The decay reaction is: $n ightarrow p + e^- + \bar{
u} + Q$.The mass defect $\Delta m$ is given by: $\Delta m = m_n - (m_p + m_e)$.Substituting the given values:$\Delta m = 1.6725 	imes 10^{-27} \ kg - (1.6725 	imes 10^{-27} \ kg + 9 	imes 10^{-31} \ kg)$.$\Delta m = -9 	imes 10^{-31} \ kg$.The magnitude of mass defect is $9 	imes 10^{-31} \ kg$.The energy released $E$ is given by $E = \Delta m c^2$,where $c = 3 	imes 10^8 \ m/s$.$E = (9 	imes 10^{-31} \ kg) 	imes (3 	imes 10^8 \ m/s)^2 = 81 	imes 10^{-15} \ J$.To convert this into $MeV$,divide by $1.6 	imes 10^{-13} \ J/MeV$:$E = \frac{81 	imes 10^{-15}}{1.6 	imes 10^{-13}} \approx 0.51 \ MeV$.

Assume that a neutron breaks into a proton and an electron. The energy released during this process is: ............ $MeV$ (mass of neutron $= 1.6725 \times 10^{-27} \ kg$,mass of proton $= 1.6725 \times 10^{-27} \ kg$,mass of electron $= 9 \times 10^{-31} \ kg$).

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