In a nuclear reaction ${ }_4^{10} \mathrm{Be}+{ }_1^1 p \rightarrow{ }_5^{10} \mathrm{~B}+{ }_0^1 n$, what is the total energy released if the binding energy of beryllium -9 is 58.2 MeV and that of boron- 10 is 64.7 MeV ?

The nuclear reaction is given by ${}_{4}^{10}\text{Be} + {}_{1}^{1}\text{p} \rightarrow {}_{5}^{10}\text{B} + {}_{0}^{1}\text{n}$.

The energy released in a nuclear reaction is given by the difference in the total binding energy of the products and the reactants. $Q = \text{BE}(\text{products}) - \text{BE}(\text{reactants})$.

The binding energy of ${}_{1}^{1}\text{p}$ (proton) is $0$ MeV and the binding energy of ${}_{0}^{1}\text{n}$ (neutron) is $0$ MeV.

The binding energy of ${}_{4}^{10}\text{Be}$ is given as $58.2$ MeV and the binding energy of ${}_{5}^{10}\text{B}$ is $64.7$ MeV.

$Q = [\text{BE}({}_{5}^{10}\text{B}) + \text{BE}({}_{0}^{1}\text{n})] - [\text{BE}({}_{4}^{10}\text{Be}) + \text{BE}({}_{1}^{1}\text{p})]$

$Q = [64.7 + 0] - [58.2 + 0] = 6.5 \text{ MeV}$.

To convert the energy from MeV to Joules, we use the conversion factor $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$.

$Q = 6.5 \times 10^{6} \times 1.602 \times 10^{-19} \text{ J} \approx 10.413 \times 10^{-13} \text{ J} = 1.0413 \times 10^{-12} \text{ J}$.

Rounding to the nearest value provided in the options, we get $1.04 \times 10^{-12} \text{ J}$.

Answer: $1.04 \times 10^{-12} J$