Question
A certain element emits $\text{K}_\alpha$ X-ray of energy 3.69keV. Use the data from the previous problem to identify the element.
✓
Answer
$\text{E}=3.69\text{Kev}=3690\text{eV}$
$\lambda=\frac{\text{hc}}{\text{E}}=\frac{1242}{3690}=0.33658\text{nm}$
$\sqrt{\frac{\text{C}}{\lambda}}\text{a}(\text{z}-\text{b})$
$\text{a}=5\times10^7\sqrt{\text{Hz}}$
$\text{b}=1.37$ (from previous problem)
$\sqrt{\frac{3\times10^8}{0.34\times10^{-9}}}$
$=5\times10^7(\text{Z}-1.37)$
$\Rightarrow\sqrt8.82\times10^{-17}=5\times10^7({\text{Z}-1.37)}$
$\Rightarrow9.39\times10^8=5\times10^7(\text{Z}-1.37)$
$\Rightarrow\frac{93.9}{5}=\text{Z}-1.37$
$\Rightarrow\text{Z}=20.15=20$
$\therefore$ The element is calcium.
$\lambda=\frac{\text{hc}}{\text{E}}=\frac{1242}{3690}=0.33658\text{nm}$
$\sqrt{\frac{\text{C}}{\lambda}}\text{a}(\text{z}-\text{b})$
$\text{a}=5\times10^7\sqrt{\text{Hz}}$
$\text{b}=1.37$ (from previous problem)
$\sqrt{\frac{3\times10^8}{0.34\times10^{-9}}}$
$=5\times10^7(\text{Z}-1.37)$
$\Rightarrow\sqrt8.82\times10^{-17}=5\times10^7({\text{Z}-1.37)}$
$\Rightarrow9.39\times10^8=5\times10^7(\text{Z}-1.37)$
$\Rightarrow\frac{93.9}{5}=\text{Z}-1.37$
$\Rightarrow\text{Z}=20.15=20$
$\therefore$ The element is calcium.
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