Radioactivity is due to

The radioactive decay of $_{35}X^{88}$ by a beta emission produces an unstable nucleus which spontaneously emits a neutron. The final product is

$_{92}U^{238}$ emits $8$ $\alpha$-particles and $6$ $\beta$-particles. The neutron/proton ratio in the product nucleus is

An atomic nucleus having a low $n/p$ ratio tries to find stability by:

In the carbon cycle,from which hot stars obtain their energy,the $_6C^{14}$ nucleus is

Let $m_p$ be the mass of a proton,$m_n$ that of a neutron,$M_1$ that of a $_{10}^{20}Ne$ nucleus,and $M_2$ that of a $_{20}^{40}Ca$ nucleus. Then: