The Ka X-ray of molybdenum has wavelength 0.071 nm.
Question:

The $\mathrm{K}_{\alpha}$ X-ray of molybdenum has wavelength $0.071 \mathrm{~nm}$. If the energy of a molybdenum atoms with a $\mathrm{K}$ electron knocked out is $27.5 \mathrm{keV}$, the energy of this atom when an $\mathrm{L}$ electron is knocked out will be $\mathrm{keV}$. (Round off to the nearest integer)

$\left[\mathrm{h}=4.14 \times 10^{-15} \mathrm{eVs}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right]$

Solution:

$\mathrm{E}_{\mathrm{k}_{\mathrm{a}}}=\mathrm{E}_{\mathrm{k}}-\mathrm{E}_{\mathrm{L}}$

$\frac{\mathrm{hc}}{\lambda_{\mathrm{k}_{a}}}=\mathrm{E}_{\mathrm{k}}-\mathrm{E}_{\mathrm{L}}$

$\mathrm{E}_{\mathrm{L}}=\mathrm{E}_{\mathrm{k}}-\frac{\mathrm{hc}}{\lambda_{\mathrm{k}_{\alpha}}}$

$=27.5 \mathrm{KeV}-\frac{12.42 \times 10^{-7} \mathrm{eVm}}{0.071 \times 10^{-9} \mathrm{~m}}$

$\mathrm{E}_{\mathrm{L}}=(27.5-17.5) \mathrm{keV}$

$=10 \mathrm{keV}$

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