The K$\alpha$ X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms 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]
Answer (integer)
10
Solution
${E_{{k_\alpha }}} = {E_k} - {E_L}$<br><br>${{hc} \over {{\lambda _{{k_\alpha }}}}} = {E_k} - {E_L}$<br><br>${E_L} = {E_k} - {{hc} \over {{\lambda _{{k_\alpha }}}}}$<br><br>= 27.5 KeV $- {{12.42 \times {{10}^{ - 7}}eVm} \over {0.071 \times {{10}^{ - 9}}m}}$<br><br>E<sub>L</sub> = (27.5 $-$ 17.5) keV<br><br>= 10 keV
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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