Q. Assume that a neutron breaks into a photon and an electron. The energy released during this process is (mass of neutron $= 1.6725 \times 10^{- 27}kg$, Mass of proton $= 1.6725 \times 10^{- 27}kg$, mass of electron $= 9 \times 10^{- 31}kg$)
A) $0.73\ MeV$
B) $7.10\ MeV$
C) $6.30\ MeV$
D) $5.4\ MeV$
Explanation
(a) $\ _{0}n^{1} \rightarrow \ _{1}H^{1} + \ _{- 1}e^{0} + \overline{v} + Q$
$\mathrm{\Delta}m = m_{n} - m_{\alpha} - m_{e}$
$= \left( 1.6725 \times 10^{- 27} - 1.6725 \times 10^{- 27} - 9 \times 10^{- 31} \right)kg$
$= - 9 \times 10^{- 31}kg$
Energy $= 9 \times 10^{- 31} \times \left( 3 \times 10^{8} \right)^{2}$
$= 0.511MeV$
Which is nearly equal to $0.73MeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.