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QJEEM21Ph4f86: JEE Main 2021 Physics - Class 12 JEEM21Ph4f86 Solution: The K X-ray of molybdenum has wavelength 0.071 nm. If the…

The K α\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]

Step-by-Step Verified Solution

Ekα=Ek−EL{E_{{k_\alpha }}} = {E_k} - {E_L}

hcλkα=Ek−EL{{hc} \over {{\lambda _{{k_\alpha }}}}} = {E_k} - {E_L}

EL=Ek−hcλkα{E_L} = {E_k} - {{hc} \over {{\lambda _{{k_\alpha }}}}}

= 27.5 KeV −12.42×10−7eVm0.071×10−9m- {{12.42 \times {{10}^{ - 7}}eVm} \over {0.071 \times {{10}^{ - 9}}m}}

EL = (27.5 −- 17.5) keV

= 10 keV

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