The $k_{α}$ $X$ -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
$(in keV)$ . (Round off to the nearest integer)
$[h = 4.14 \times 10^{- 15} eVs, c = 3 \times 10^{8} {ms}^{- 1}]$

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Solution

The total energy of $k_{α}$ can be given as
$E_{k_{α}} = E_{k} - E_{L}$
or, $\frac{h c}{λ_{k_{α}}} = E_{k} - E_{L}$
or, $E_{L} = E_{K} - \frac{h c}{λ_{k_{α}}}$
$= 27.5 keV - \frac{(4.14 \times 10^{- 15} eVs) \times (3 \times 10^{8}) {ms}^{- 1}}{(0.071 \times 10^{- 9}) m}$
$E_{L} = 27.5 keV - \frac{(12.42 \times 10^{- 7}) eVm}{(0.071 \times 10^{- 9}) m}$
or, $E_{L} = (27.5 - 17.5) keV = 10 keV$

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The kα 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 (in keV). (Round off to the nearest integer) [h=4.14×10−15 eVs, c=3×108 ms−1]

The total energy of kα can be given as Ekα=Ek−EL or, hcλkα=Ek−EL or, EL=EK−hcλkα =27.5 keV−(4.14×10−15 eVs)×(3×108) ms−1(0.071×10−9) m EL=27.5 keV−(12.42×10−7) eVm(0.071×10−9) m or, EL=(27.5−17.5) keV=10 keV