The energy equivalent of $1.0 \text{ kg}$ of substance is . . . . . . .

We are given the following atomic masses:
$^{238}_{92}U = 238.05079 \; u$
$^{4}_{2}He = 4.00260 \; u$
$^{234}_{90}Th = 234.04363 \; u$
$^{1}_{1}H = 1.00783 \; u$
$^{237}_{91}Pa = 237.05121 \; u$
Here,the symbol $Pa$ represents the element protactinium $(Z=91)$.
$(a)$ Calculate the energy released during the alpha decay of $^{238}_{92}U$.
$(b)$ Show that $^{238}_{92}U$ cannot spontaneously emit a proton.

$A$ nucleus of mass $M$ emits a $\gamma$-ray photon of frequency $\nu$. The loss of internal energy by the nucleus is:

The binding energy per nucleon for a deuteron and an $\alpha$-particle are $x_1$ and $x_2$ respectively. The energy $Q$ released in the following reaction is:
$_1H^2 + _1H^2 \rightarrow {_2}{He}^4 + Q$

If the binding energy per nucleon of a nuclide is high,then:

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