Suppose a \({}^{40}_{20}\mathrm{Ca}\) nucleus splits into two equal fragments, \({}^{20}_{10}\mathrm{Ne}\). Given: \(m({}^{40}_{20}\mathrm{Ca}) = 39.96259\ \mathrm{u}\), \(m({}^{20}_{10}\mathrm{Ne}) = 19.99244\ \mathrm{u}\) and \(1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^{2}\). The Q-value of this process is about
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-20.8\ \mathrm{MeV}\)
The Q-value of a nuclear process is \( Q=(\text{sum of initial masses}-\text{sum of final masses})c^{2} \). Both sides have 20 electrons, so the atomic masses can be used as given.
\( \begin{aligned} \Delta m &= m(\mathrm{Ca})-2\,m(\mathrm{Ne}) \\ &= 39.96259-2\times19.99244 \\ &= -0.02229\ \mathrm{u} \end{aligned} \)
\( \begin{aligned} Q &= \Delta m\,c^{2} \\ &= -0.02229\times931.5\ \mathrm{MeV} \\ &\approx -20.8\ \mathrm{MeV} \end{aligned} \)
From the above, \( Q \) is negative. Hence, this splitting cannot happen on its own. About 20.8 MeV would have to be supplied to make it happen.
NCERT (2026-27):
Summary – “9. The Q-value of a nuclear process is … Due to conservation of mass-energy, this is also, Q = (sum of initial masses – sum of final masses)c2”