Given mass of electron (m) = 5.486 x 10⁻⁴ a.m.u.

Given mass of electron ($m$) = $5.486 \times 10^{-4} \, \text{a.m.u.}$
The standard conversion factor for a.m.u. to kg is $1 \, \text{a.m.u.} = 1.660539 \times 10^{-27} \, \text{kg}$.

$$m = (5.486 \times 10^{-4} \, \text{a.m.u.}) \times (1.660539 \times 10^{-27} \, \text{kg/a.m.u.})$$ $$m = 9.10938 \times 10^{-31} \, \text{kg}$$

The formula is $E = mc^2$, where:

• $E$ is the energy
• $m$ is the mass in kilograms
• $c$ is the speed of light in a vacuum ($3.0 \times 10^8 \, \text{m/s}$)

Substitute the values:

$$E = (9.10938 \times 10^{-31} \, \text{kg}) \times (3.0 \times 10^8 \, \text{m/s})^2$$ $$E = (9.10938 \times 10^{-31}) \times (9.0 \times 10^{16}) \, \text{J}$$ $$E = (9.10938 \times 9.0) \times (10^{-31} \times 10^{16}) \, \text{J}$$ $$E = 81.98442 \times 10^{-15} \, \text{J}$$ $$E = 8.198442 \times 10^{-14} \, \text{J}$$

Given conversion factor: $1 \, \text{MeV} = 1.602 \times 10^{-13} \, \text{J}$.
To convert Joules to MeV, divide by the conversion factor:

$$E_{\text{MeV}} = \frac{E_{\text{J}}}{1.602 \times 10^{-13} \, \text{J/MeV}}$$ $$E_{\text{MeV}} = \frac{8.198442 \times 10^{-14} \, \text{J}}{1.602 \times 10^{-13} \, \text{J/MeV}}$$ $$E_{\text{MeV}} = \frac{8.198442}{1.602} \times \frac{10^{-14}}{10^{-13}} \, \text{MeV}$$ $$E_{\text{MeV}} = 5.117629 \times 10^{-1} \, \text{MeV}$$ $$E_{\text{MeV}} = 0.5117629 \, \text{MeV}$$

Rounding to three significant figures (consistent with the given mass and speed of light):

$$E_{\text{MeV}} \approx 0.512 \, \text{MeV}$$

The energy of the electron is $\text{0.512 MeV}$.