JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Modern Physics

The

K_{α}

-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 ____ ke V. (Round off to the nearest integer)

[h = 4.14 \times 1 0^{- 15} eVs, c = 3 \times 1 0^{8} ms^{- 1}]

Answer: 10

Solution

The

K_{α}

photon is emitted when an L electron drops into the vacancy in the K shell, so its energy is the difference between the energy of the atom with a K vacancy and the energy of the atom with an L vacancy:

E_{K_{α}} = \frac{h c}{λ} = \frac{( 4.14 \times 1 0 ^{- 15} ) ( 3 \times 1 0 ^{8} )}{0.071 \times 1 0 ^{- 9}} = 1.75 \times 1 0^{4} eV = 17.5 keV

E_{L} = E_{K} - E_{K_{α}} = 27.5 - 17.5 = 10 keV

The energy of the atom with an L electron knocked out is

10

keV.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer

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JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Modern Physics

Solution

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