If the mass of the iron nucleus is \(55.85~\text{u}\) and \(\mathrm{A} = 56\), the nuclear density of the iron is:

The energy equivalent of \(1\) g of substance is:
1. \(8.3\times10^{13}~\text{J}\)
2. \(9\times10^{13}~\text{J}\)
3. \(7.7\times10^{13}~\text{J}\)
4. \(11\times10^{13}~\text{J}\)

We are given the following atomic masses:

${}_{92}{}^{238}\mathrm{U}$ = 238.05079 u, ${}_2{}^4\mathrm{He}$= 4.00260 u

${}_{90}{}^{234}\mathrm{Th}$ = 234.04363 u, ${}_1{}^1\mathrm{H}$= 1.00783 u

${}_{91}{}^{237}\mathrm{Pa}$= 237.05121 u

Here the symbol Pa is for the element protactinium (Z = 91).

The energy released during the alpha decay of ${}_{92}{}^{238}\mathrm{U}$ is:

1. 6.14 MeV 

2. 7.68 MeV 

3. 4.25 MeV

4. 5.01 MeV 

We are given the following atomic masses:

${}_{92}{}^{238}\mathrm{U}$ = 238.05079 u, ${}_2{}^4\mathrm{He}$= 4.00260 u

${}_{90}{}^{234}\mathrm{Th}$ = 234.04363 u, ${}_1{}^1\mathrm{H}$= 1.00783 u

${}_{91}{}^{237}\mathrm{Pa}$= 237.05121 u

Here the symbol Pa is for the element protactinium (Z = 91).

Then:

1. ${}_{92}{}^{238}\mathrm{U}$ can not spontaneously emit a proton.

2. ${}_{92}{}^{238}\mathrm{U}$ can spontaneously emit a proton.

3. The Q-value of the process is negative.

4. Both (1) and (3)