(C) The wavelength of the probe is given by the de Broglie relation $\lambda = h/p$. For a structure of size $\lambda \approx 10^{-15} \;m$,the momentum $p$ is $p = h/\lambda = (6.63 \times 10^{-34} \;J \cdot s) / (10^{-15} \;m) = 6.63 \times 10^{-19} \;kg \cdot m/s$.
Since the energy of the electron is very high,we use the relativistic energy-momentum relation $E^2 = p^2c^2 + m_0^2c^4$. Given that $pc \approx (6.63 \times 10^{-19} \;kg \cdot m/s) \times (3 \times 10^8 \;m/s) \approx 1.99 \times 10^{-10} \;J$,which is much greater than the rest mass energy $m_0c^2 = 0.511 \;MeV \approx 8.19 \times 10^{-14} \;J$,we can approximate $E \approx pc$.
Thus,$E \approx 1.99 \times 10^{-10} \;J$.
Converting this to electron-volts: $E = (1.99 \times 10^{-10} \;J) / (1.6 \times 10^{-19} \;J/eV) \approx 1.24 \times 10^9 \;eV = 1.24 \;GeV$.
Therefore,the order of energy of these electron beams is approximately $1 \;GeV$.