In the nuclear process n to p + e^- + bar{nu} | vedclass

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Question

(B) The nuclear process is $n 	o p + e^- + \bar{
u}$.Mass of reactant (neutron) $m_n = 1.6747 	imes 10^{-27} \ kg$.Mass of products (proton + elect

Vedclass · Accepted Answer

$0.73$

Vedclass · Answer

(B) The nuclear process is $n 	o p + e^- + \bar{
u}$.Mass of reactant (neutron) $m_n = 1.6747 	imes 10^{-27} \ kg$.Mass of products (proton + electron) $m_p + m_e = (1.6725 	imes 10^{-27} + 9 	imes 10^{-31}) \ kg$.$m_p + m_e = (1.6725 + 0.0009) 	imes 10^{-27} \ kg = 1.6734 	imes 10^{-27} \ kg$.The mass defect $\Delta m = m_n - (m_p + m_e) = (1.6747 - 1.6734) 	imes 10^{-27} \ kg = 0.0013 	imes 10^{-27} \ kg = 1.3 	imes 10^{-30} \ kg$.The energy released $E = \Delta m c^2 = (1.3 	imes 10^{-30}) 	imes (3 	imes 10^8)^2 \ J$.$E = 1.3 	imes 10^{-30} 	imes 9 	imes 10^{16} \ J = 11.7 	imes 10^{-14} \ J$.To convert energy into $MeV$,divide by $1.6 	imes 10^{-13} \ J/MeV$:$E = \frac{11.7 	imes 10^{-14}}{1.6 	imes 10^{-13}} \ MeV = \frac{1.17}{1.6} \ MeV \approx 0.73 \ MeV$.

In the nuclear process $n \to p + e^- + \bar{\nu}$,if the masses of proton,neutron,and electron are $1.6725 \times 10^{-27} \ kg$,$1.6747 \times 10^{-27} \ kg$,and $9 \times 10^{-31} \ kg$ respectively,then the energy released is ...... $MeV$.

Similar Questions

$A$ nuclide $1$ is said to be the mirror isobar of nuclide $2$ if $Z_1 = N_2$ and $Z_2 = N_1$. $(a)$ What nuclide is a mirror isobar of $_{11}^{23}Na$? $(b)$ Which nuclide out of the two mirror isobars has greater binding energy and why?

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