The K α X-ray of molybdenum has wavelength 0.071 nm. If the energy of… | JEE Main 2021 (27 Jul, Shift 2)

Q5JEE Main 2021 · 27 JUL · SHIFT 2Medium

Quantum Numbers, X-rays & Other Questions

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 $\times$ 108 ms ${}^{-1}$ ]

Numerical answer type. Work it out, then open the answer.

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Answer: 10

$K_\alpha$ photon energy $E = \frac{hc}{\lambda} = \frac{4.14\times10^{-15}\times 3\times10^{8}}{0.071\times10^{-9}}$
$E \approx 1.75\times 10^{4}$ eV $= 17.5$ keV
$K_\alpha$ is emitted when an L electron fills the K hole, so $E_L = E_K - E_{K_\alpha} = 27.5 - 17.5$
Answer: 10 keV

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