The nucleus of a radioactive element possesses:

Radioactivity is due to

$_{92}U^{235} + n \to \text{fission product} + \text{neutron} + 3.20 \times 10^{-11} \ J$. The energy released when $1 \ g$ of $_{92}U^{235}$ undergoes fission 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:

The atomic mass of an element is $12.00710 \, amu$. If there are $6$ neutrons in the nucleus of the atom of the element,the binding energy per nucleon of the nucleus will be $....... \, MeV$.
($e^{-} = 0.00055 \, amu$,$p = 1.00814 \, amu$,$n = 1.00893 \, amu$)

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