The $$K_\alpha$$ X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a $$K$$ electron knocked out is 27.5 keV, the energy of this atom when an $$L$$ electron is knocked out will be _________ keV. (Round off to the nearest integer) [$$h = 4.14 \times 10^{-15}$$ eV s, $$c = 3 \times 10^8$$ m s$$^{-1}$$]

We are told that the characteristic $$K_\alpha$$ line of molybdenum has wavelength $$\lambda = 0.071 \text{ nm}$$ and that the energy required to knock out a $$K$$ electron (that is, the $$K$$-shell binding energy) is $$E_K = 27.5 \text{ keV}$$. Our task is to find the energy needed to knock out an $$L$$ electron, which we shall call $$E_L$$.

First we convert the given wavelength to metres so that we can work with SI units:

$$\lambda \;=\; 0.071$$ nm $$= 0.071 \times 10^{-9}$$ m $$= 7.1 \times 10^{-11}$$ m $$.$$

Now we calculate the energy of the emitted $$K_\alpha$$ photon. The photon-energy formula is stated as

$$E_{\text{photon}} \;=\; \dfrac{h\,c}{\lambda},$$

where $$h = 4.14 \times 10^{-15} \text{ eV·s}$$ and $$c = 3 \times 10^{8} \text{ m·s}^{-1}$$.

Substituting the values, we obtain

$$ \begin{aligned} E_{\text{photon}} &= \dfrac{\bigl(4.14 \times 10^{-15}\ \text{eV·s}\bigr)\,\bigl(3 \times 10^{8}\ \text{m·s}^{-1}\bigr)} {7.1 \times 10^{-11}\ \text{m}} \\[4pt] &= \dfrac{12.42 \times 10^{-7}\ \text{eV·m}}{7.1 \times 10^{-11}\ \text{m}} \\[4pt] &= \left(\dfrac{12.42}{7.1}\right)\times 10^{\,(-7+11)}\ \text{eV} \\[4pt] &= 1.75 \times 10^{4}\ \text{eV} \\[4pt] &= 17.5 \text{ keV}. \end{aligned} $$

In a $$K_\alpha$$ transition an electron falls from the $$L$$ shell to the $$K$$ shell. Therefore the energy of the photon equals the difference between the two binding energies:

$$E_{\text{photon}} = E_K - E_L.$$ Rearranging this gives

$$E_L = E_K - E_{\text{photon}}.$$

Substituting the numerical values, we have

$$ \begin{aligned} E_L &= 27.5 \text{ keV} - 17.5 \text{ keV} \\[4pt] &= 10.0 \text{ keV}. \end{aligned} $$

Rounded to the nearest integer, the energy required to knock out an $$L$$ electron is $$10 \text{ keV}$$.

So, the answer is $$10$$.