Question

The $K_a\text{X-ray}$ of aluminium $(Z=13)$ and zinc $(Z=30)$ have wavelengths $887\text{pm}$ and $146\text{pm}$ respectively. Use Moseley’s law $\sqrt{\nu }=a(Z-b)$ to find the wavelength of the $K_a\text{X-ray}$ of iron $(Z=26)$.$(hc=1242\text{eV-nm})$

• A. $298\text{pm}$
• B. $398\text{pm}$
• C. $499\text{pm}$
• D. $198\text{pm}$

Answer 1

The correct option is D $198\text{pm}$

For, ${\lambda }_1=887\text{pm}$

${\nu }_1=\frac{c}{\lambda }=\frac{3\times {10}^8}{887\times {10}^{-12}}$

$=3.382\times {10}^{17}=33.82\times {10}^{16}\text{Hz}$

$\therefore \sqrt{{\nu }_1}=5.815\times {10}^8$

For, ${\lambda }_2=146\text{pm}$

${\nu }_2=\frac{3\times {10}^8}{146\times {10}^{-12}}=0.02054\times {10}^{20}=2.054\times {10}^{18}\text{Hz}$

$\therefore \sqrt{{\nu }_2}=1.4331\times {10}^9$

Using $\sqrt{\nu }=a(Z-b)$ for aluminium and zinc, we have

$\Rightarrow \frac{5.815\times {10}^8}{1.4331\times {10}^9}=\frac{a(13-b)}{a(30-b)}$

$\Rightarrow \frac{13-b}{30-b}=\frac{5.815\times {10}^{-1}}{1.4331}=0.4057$

$\Rightarrow 30\times 0.4057-0.4057b=13-b$

$\Rightarrow 12.171-\mathrm{0.4.57}b+b=13$

$\Rightarrow b=\frac{0.829}{0.5943}=1.39491$

$\Rightarrow a=\frac{5.815\times {10}^8}{11.33}=0.51323\times {10}^8=5\times {10}^7$

For iron:

$\sqrt{\nu }=5\times {10}^7(26-1.39)=5\times 24.61\times {10}^7=123.05\times {10}^7$

$\therefore \nu =15141.3\times {10}^{14}\text{Hz}$

Now,

$\lambda =\frac{c}{\nu }=\frac{3\times {10}^8}{15141.3\times {10}^{14}}$

$=0.000198\times {10}^{-6}\text{m}=198\times {10}^{-12}\text{m}=198\text{pm}$

Hence, $\left(D\right)$ is the correct answer.