The nucleus \( {}^{23}_{10}\mathrm{Ne} \) decays by \( \beta^{-} \) emission. The maximum kinetic energy of the electrons emitted is:
Given that:
\( m({}^{23}_{10}\mathrm{Ne}) = 22.994466\,\mathrm{u} \), \( m({}^{23}_{11}\mathrm{Na}) = 22.989770\,\mathrm{u} \) and \( 1\,\mathrm{u} = 931\,\mathrm{MeV}/c^{2} \)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4.372 MeV
In \( \beta^{-} \) decay a neutron changes into a proton and the nucleus emits an electron and an antineutrino. The energy released is the Q-value, the mass defect times \( c^{2} \), and it is shared by the electron, the antineutrino and the recoiling nucleus.
\( \displaystyle {}^{23}_{10}\mathrm{Ne} \rightarrow {}^{23}_{11}\mathrm{Na} + e^{-} + \bar{\nu} + Q \)
The maximum kinetic energy of the electron is obtained when the antineutrino carries away almost no energy and the heavy daughter nucleus takes a negligible recoil energy, so \( K_{\max} \approx Q \).
Using atomic masses, the mass of the emitted electron is accounted for by the extra electron in the neutral Na atom, so the electron masses cancel:
\( \displaystyle Q = \left[m(\mathrm{Ne}) - m(\mathrm{Na})\right]c^{2} \)
\( \displaystyle \Delta m = 22.994466 - 22.989770 = 0.004696\,\mathrm{u} \)
\( \displaystyle K_{\max} = 0.004696 \times 931\,\mathrm{MeV} = 4.372\,\mathrm{MeV} \)
Note: It seems that there is a mistake in the original NTA paper, which prints ‘nucleus’ masses when atomic masses are meant: the given values are the masses of the neutral Ne and Na atoms. If the given values were truly nuclear masses, the emitted electron's mass would have to be subtracted, \( Q = (\Delta m - m_e)c^{2} = (0.004696 - 0.000549) \times 931 = 3.861\,\mathrm{MeV} \), which is not among the options.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Exercises – “… 13.14 The nucleus \( {}^{23}_{10}\mathrm{Ne} \) decays by β− emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted.”