Question

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 = $4.14 \times 10^{-15}$ eVs, c = $3 \times 10^8$ ms$^{-1}$ ]

Hint:

The notation K$_\alpha$ refers to the X-ray emitted when an electron transitions from the L-shell (n=2) to the K-shell (n=1). Similarly, K$_\beta$ is from M-shell (n=3) to K-shell (n=1). The energy of the emitted photon is the difference in the binding energies of these shells.

Answer 1

Correct Answer: 10

The energy of a K$_\alpha$ X-ray photon ($E_{K\alpha}$) corresponds to the energy difference between an atom with a vacancy in the K-shell (n=1) and an atom with a vacancy in the L-shell (n=2).
$E_{K\alpha} = E_{K-vacancy} - E_{L-vacancy}$.
First, let's calculate the energy of the K$_\alpha$ photon from its wavelength. We can use the formula $E = \frac{hc}{\lambda}$.
A useful shortcut is $E(\text{eV}) = \frac{1240 \text{ eV}\cdot\text{nm}}{\lambda(\text{nm})}$.
Given $\lambda_{K\alpha} = 0.071$ nm.
$E_{K\alpha} = \frac{1240}{0.071} \approx 17465 \text{ eV} = 17.465 \text{ keV}$.
We are given that the energy of the atom with a K electron knocked out is $E_{K-vacancy} = 27.5$ keV. This is the binding energy of the K-shell electron.
The question asks for the energy of the atom when an L electron is knocked out, which is $E_{L-vacancy}$.
Rearranging the formula: $E_{L-vacancy} = E_{K-vacancy} - E_{K\alpha}$.
$E_{L-vacancy} = 27.5 \text{ keV} - 17.465 \text{ keV} = 10.035 \text{ keV}$.
The question asks to round off to the nearest integer.
Rounding 10.035 to the nearest integer gives 10.
So, the energy of the atom with an L electron knocked out is 10 keV.