Given data

• Wavelength of $K_{\alpha}$ X-ray: $\lambda = 0.071\ \text{nm} = 0.071 \times 10^{-9}\ \text{m}$
• Energy of atom with a K-electron knocked out: $E_K = 27.5\ \text{keV}$
• Constants: $h = 4.14 \times 10^{-15}\ \text{eV s}, \qquad c = 3 \times 10^8\ \text{m/s}$

Meaning of $K_{\alpha}$ X-ray

A $K_{\alpha}$ X-ray is emitted when an electron falls from the $L$-shell to the $K$-shell.

Hence, the photon energy is: $E_{K\alpha} = E_K - E_L$ where:

• $E_K$ = binding energy of K-shell electron
• $E_L$ = binding energy of L-shell electron

Calculate the energy of the emitted $K_{\alpha}$ photon

Using $E = \frac{hc}{\lambda}$

First compute $hc$: $hc = (4.14 \times 10^{-15})(3 \times 10^8) = 12.42 \times 10^{-7}\ \text{eV m}$

Now, $E_{K\alpha} = \frac{12.42 \times 10^{-7}}{0.071 \times 10^{-9}}\ \text{eV}$

$= \frac{12.42}{0.071} \times 10^2\ \text{eV}$

$\approx 174.93 \times 10^2\ \text{eV}$

$\approx 1.7493 \times 10^4\ \text{eV} = 17.5\ \text{keV}$

Find the L-shell binding energy

Since $E_{K\alpha} = E_K - E_L$ we get $E_L = E_K - E_{K\alpha}$

$E_L = 27.5 - 17.5 = 10.0\ \text{keV}$

Final answer

The energy of the atom when an $L$ electron is knocked out is: $\boxed{10\ \text{keV}}$

Rounded to the nearest integer: $\boxed{10}$