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

To determine the energy of a molybdenum atom when an L electron is knocked out, we begin by calculating the energy corresponding to the given K_α X-ray wavelength. The energy E of a photon is given by the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength.
Given:

• λ = 0.071 nm = 0.071 × 10^-9 m
• h = 4.14 × 10^-15 eVs
• c = 3 × 10⁸ ms^-1

Substituting these values, we get:
E = (4.14 × 10^-15 eVs)(3 × 10⁸ ms^-1) / (0.071 × 10^-9 m) = 17.48 keV.
When a K electron is knocked out, the energy of the atom is 27.5 keV. The K_α X-ray transition corresponds to the energy difference between the K and L shell. Therefore, the energy of the atom when the L electron is knocked out is:
E_L = E_K - E_Kα = 27.5 keV - 17.48 keV = 10.02 keV.
Rounding 10.02 keV to the nearest integer, we obtain 10 keV.
This result falls within the specified range of 10,10. Therefore, the energy of the atom with an L electron knocked out is 10 keV.