Question

The K_α X-rays of aluminium (Z = 13) and zinc (Z = 30) have wavelengths 887 pm and 146 pm respectively. Use Moseley's law √v = a(Z − b) to find the wavelengths of the K_α X-ray of iron (Z = 26).

Answer 1

Given:
Wavelength of K_α X-rays of aluminium, λ₁ = 887 pm
Frequency of X-rays of aluminium is given by
$$\begin{gathered} {\nu }_a=\frac{c}{\lambda } \\ {\nu }_a=\frac{3\times {10}^8}{887\times {10}^{-12}} \\ {\nu }_a=3.382\times {10}^{17} \\ {\nu }_a=33.82\times {10}^{16}\mathrm{Hz} \end{gathered}$$

Wavelength of K_α X-rays of zinc, ${\lambda }_2$ = 146 pm
Frequency of X-rays of zinc is given by
$$\begin{gathered} {\nu }_z=\frac{3\times {10}^8}{146\times {10}^{-12}} \\ {\nu }_z=0.02055\times {10}^{20} \\ {\nu }_z=2.055\times {10}^{18}\mathrm{Hz} \end{gathered}$$

We know
$\sqrt{\nu }$ = a(Z − b)

For aluminium,
5.815 × 10⁸ = a(13 − b) ...(1)

For zinc,
1.4331 × 10⁹ = a(30 − b) ...(2)

Dividing (1) by (2)
$$\begin{gathered} \frac{13-b}{30-b}=\frac{5.815\times {10}^{-1}}{1.4331} \\ =0.4057 \\ \Rightarrow 30\times 0.4057-0.4057b=13-b \\ \Rightarrow 12.171-0.4057b+b=13 \\ b=\frac{0.829}{0.5943}=1.39491 \\ a=\frac{5.815\times {10}^8}{11.33} \\ =0.51323\times {10}^8=5\times {10}^7 \end{gathered}$$

For Fe,
Frequency $\left(\nu \text{'}\right)$ is given by
$\nu$' = 5× 10⁷ (26 − 1.39)
= 5 × 24.61 × 10⁷
= 123.05 × 10⁷

$\nu$' = $\frac{c}{\lambda }$
Here, c = speed of light
$\lambda$ = Wavelength of light
$$\begin{gathered} \therefore \frac{c}{\lambda }=5141.3\times {10}^{14} \\ \Rightarrow \lambda =\frac{3\times {10}^8}{5141.3\times {10}^{14}} \\ =0.000198\times {10}^{-5}\mathrm{m} \\ =198\times {10}^{-12}=198\mathrm{pm} \end{gathered}$$