Problem: The easiest fusion reaction to initiate is21H + 31H → 42He + 10nCalculate the energy released per  42He nucleus produced and per mole of  42He produced. The atomic masses are 21H, 2.01410 u; 31H, 3.01605 u; and 42He, 4.00260 u. The masses of the electron and neutron are 5.4858 X 10-4 u and 1.00866 u, respectively.

FREE Expert Solution

To calculate the energy released:

21H + 31H → 42He + 10n

Step 1: mass defect (Δm)

$\overline{){\mathbf{∆}}{\mathbf{m}}{\mathbf{=}}{\mathbf{p}}{\mathbf{r}}{\mathbf{o}}{\mathbf{d}}{\mathbf{u}}{\mathbf{c}}{\mathbf{t}}{\mathbf{s}}{\mathbf{-}}{\mathbf{r}}{\mathbf{e}}{\mathbf{a}}{\mathbf{c}}{\mathbf{t}}{\mathbf{a}}{\mathbf{n}}{\mathbf{t}}{\mathbf{s}}}$

Δm = – 0.01889 u or amu

Convert to kg: 1 amu = 1.6606x10-27 kg

Δm = – 3.1369 x10-29 kg

Problem Details

The easiest fusion reaction to initiate is

21H + 31H → 42He + 10n

Calculate the energy released per  42He nucleus produced and per mole of  42He produced. The atomic masses are 21H, 2.01410 u; 31H, 3.01605 u; and 42He, 4.00260 u. The masses of the electron and neutron are 5.4858 X 10-4 u and 1.00866 u, respectively.