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JEE Main 202127 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Evening Shift
Subject: Physics
Chapter: Atomic Physics

JEE Main 2021 · 27 July · Evening Shift · Physics · Atomic Physics

The K _ α X-ray of molybdenum has wavelength 0 . 071nm . If the...

The

{K}_{\alpha}

X-ray of molybdenum has wavelength

0 . 071nm

. If the energy of a molybdenum atom with a

K

electron knocked out is

27 . 5keV

, the energy of this atom when an

L

electron is knocked out will be

keV

. (Round off to the nearest integer )

\left[h = 4 . 14 \times {10}^{- 15}eVs, c = 3 \times {10}^{8}{ms}^{- 1}\right]

Correct answer

0

Step-by-step solution

${E}_{{k}_{\alpha}} = {E}_{k} - {E}_{L}$

$\dfrac{hc}{{\lambda}_{{k}_{\alpha}}} = {E}_{k} - {E}_{L}$

${E}_{L} = {E}_{k} - \dfrac{hc}{{\lambda}_{{k}_{\alpha}}}$

$= 27 . 5KeV - \dfrac{12 . 42 \times {10}^{- 7}eVm}{0 . 071 \times {10}^{- 9}m}$

${E}_{L} = ( 27 . 5 - 17 . 5 ) keV$

$= 10keV$

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