First slide

X-rays

Question

If the $K_{α}$ radiation of Mo (Z = 42) has a wavelength of 0.71 $Å$ the wavelength of the corresponding radiation of Cu (Z = 29)

Moderate

A

1 $Å$
B

2 $Å$
C

1.52 $Å$
D

1.25 $Å$

Solution

Moseley's law
$F o r K_{α}, λ \propto \frac{1}{{(Z - 1)}^{2}}$
$\frac{λ_{2}}{λ_{1}} = \frac{{(Z_{1} - 1)}^{2}}{{(Z_{2} - 1)}^{2}} \Rightarrow \frac{λ_{2}}{0.71} = \frac{(42 - 1)^{2}}{(29 - 1)^{2}} \Rightarrow λ_{2} = 1.522 \dot{A}$

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