The correct statement of the following 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.

$A$ radioactive nucleus has specific binding energy $E_{1}$. It emits an $\alpha$-particle. The resulting nucleus has specific binding energy $E_{2}$. Then

The energy equivalent to a mass of $1 \,kg$ is

$A$ nucleus of mass $ 20 u $ emits a $ \gamma $ photon of energy $ 6 MeV $. If the emission is assumed to occur when the nucleus is free and at rest, then the nucleus will have a kinetic energy nearest to (take $ 1 u = 1.6 \times 10^{-27} kg $): (in $keV$)

If the speed of light were $2/3$ of its present value,the energy released in a given atomic explosion will be decreased by a fraction: