The mass of a $H$-atom is less than the sum of the masses of a proton and an electron. Why is this?
(N/A) When a proton and an electron combine to form a $H$-atom,the system moves to a state of lower potential energy to attain maximum stability. According to Einstein's mass-energy equivalence principle,$E = mc^2$,this reduction in potential energy corresponds to a decrease in the mass of the system. This mass defect is released as energy (binding energy) during the formation of the atom. Consequently,the mass of the $H$-atom is less than the sum of the masses of its constituent particles in their free state.
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The binding energy for the following nuclear reactions are expressed in $MeV$.
${ }_2 He ^3+{ }_0 n ^1 \rightarrow{ }_2 He ^4+20 \ MeV$
${ }_2 He ^4+{ }_0 n ^1 \rightarrow{ }_2 He ^5-0.9 \ MeV$
If $X_3, X_4, X_5$ denote the stability of ${ }_2 He ^3, { }_2 He ^4$ and ${ }_2 He ^5$, respectively, then the correct order is:
${ }_2 He ^3+{ }_0 n ^1 \rightarrow{ }_2 He ^4+20 \ MeV$
${ }_2 He ^4+{ }_0 n ^1 \rightarrow{ }_2 He ^5-0.9 \ MeV$
If $X_3, X_4, X_5$ denote the stability of ${ }_2 He ^3, { }_2 He ^4$ and ${ }_2 He ^5$, respectively, then the correct order is:
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