A radioactive nucleus of mass number A and atomic number Z, emits an alpha particle first and then a beta negative particle. The mass number and atomic number of the daughter nucleus will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( A-4,\ Z-1 \)
An alpha particle is a helium nucleus, \( {}^{4}_{2}\mathrm{He} \), so alpha emission lowers \( A \) by 4 and \( Z \) by 2. A beta-negative particle is an electron created in the nucleus when a neutron changes into a proton, so beta emission leaves \( A \) unchanged and raises \( Z \) by 1.
\( \displaystyle {}^{A}_{Z}\mathrm{X} \rightarrow {}^{A-4}_{Z-2}\mathrm{Y} + {}^{4}_{2}\mathrm{He} \)
\( \displaystyle {}^{A-4}_{Z-2}\mathrm{Y} \rightarrow {}^{A-4}_{Z-1}\mathrm{W} + {}^{\,0}_{-1}\mathrm{e} + \bar{\nu} \)
The daughter nucleus has mass number \( A-4 \) and atomic number \( Z-1 \).
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):
Alpha Decay – “… In this process, it is observed that since \( {}^{4}_{2}\mathrm{He} \) contains two protons and two neutrons, the mass number and the atomic number of the daughter nucleus decreases by four and two, respectively.”
Beta Decay – “… In beta-minus decay, a neutron transforms into a proton within the nucleus according to
\( \displaystyle n \rightarrow p + e^{-} + \bar{\nu} \qquad (13.24) \)
whereas in beta-plus decay, a proton transforms into neutron (inside the nucleus) via
\( \displaystyle p \rightarrow n + e^{+} + \nu \qquad (13.25) \)
These processes show why the mass number \( A \) of a nuclide undergoing beta decay does not change; one of its constituent nucleons simply changes its character according to Eq. (13.24) or (13.25).”