Question 400: Atoms & Nuclei

Question

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 $$	imes$$ 10$$-$$15 eVs, c = 3 $$	imes$$ 108 ms$$-$$1]

Accepted Answer

$${E_{{k_\alpha }}} = {E_k} - {E_L}$$
$${{hc} \over {{\lambda _{{k_\alpha }}}}} = {E_k} - {E_L}$$
$${E_L} = {E_k} - {{hc} \over {{\lambda _{{k_\alpha }}}}}$$
= 27.5 KeV $$ - {{12.42 	imes {{10}^{ - 7}}eVm} \over {0.071 	imes {{10}^{ - 9}}m}}$$
E
L
= (27.5 $$-$$ 17.5) keV
= 10 keV

Atoms & Nuclei

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