Interpret the nuclear composition condition

For the given unstable nucleus, $\frac{N}{P} < 1$ This means the nucleus is proton-rich (or neutron-deficient), since the number of neutrons is less than the number of protons.

Recall how different decay modes change $N$ and $P$

We examine each option by seeing whether it helps a proton-rich nucleus move toward stability.

(A) $\beta^-$ decay

In $\beta^-$ decay: $n \to p + e^- + \bar\nu$ So,

• $N$ decreases by 1
• $P$ increases by 1

Hence $N/P$ decreases even further. A proton-rich nucleus does not prefer this mode.

Therefore, A is incorrect.

(B) Orbital or K-electron capture

In electron capture: $p + e^- \to n + \nu$ So,

• $P$ decreases by 1
• $N$ increases by 1

This increases the $N/P$ ratio and helps a proton-rich nucleus approach stability.

Therefore, B is correct.

(C) Neutron emission

In neutron emission:

• $N$ decreases by 1
• $P$ unchanged

This makes the nucleus even more neutron-deficient, so it moves away from stability when $N/P < 1$.

Therefore, C is incorrect.

(D) $\beta^+$ decay (positron emission)

In $\beta^+$ decay: $p \to n + e^+ + \nu$ So,

• $P$ decreases by 1
• $N$ increases by 1

Again, this raises $N/P$ and helps the proton-rich nucleus move toward stability.

Therefore, D is correct.

Final selection

The possible decay modes for a nucleus with $\frac{N}{P} < 1$ are: $\boxed{\text{B, D}}$

Comparison with stored correct answer

Stored correct answer: B, D

Our derived answer: B, D

Hence, the derived answer agrees with the stored answer.