Question

A certain element emits $K_{\alpha }\text{X-ray}$ of energy $3.69\text{keV}$. The $K_{\alpha }\text{X-ray}$ 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 element.$(hc=1242\text{eV-nm})$

• A. $\text{Sodium}$
• B. $\text{Molybdenum}$
  
• C. $\text{Calcium}$
• D. $\text{Argon}$

Answer 1

The correct option is C $\text{Calcium}$

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

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

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

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

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

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

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

Given, $E=3.69\text{keV}=3690\text{eV}$

$\Rightarrow \lambda =\frac{hc}{E}=\frac{1242\text{eV-nm}}{3690\text{eV}}=0.33658\text{nm}$

Now,

$\sqrt{\frac{c}{\lambda }}=a(Z-b)$

$\Rightarrow \sqrt{\frac{3\times {10}^8}{0.34\times {10}^{-9}}}=5\times {10}^7(Z-1.37)$

$\Rightarrow \sqrt{8.82\times {10}^{17}}=5\times {10}^7(Z-1.37)$

$\Rightarrow 9.39\times {10}^8=5\times {10}^7(Z-1.37)$

$\Rightarrow \frac{93.9}{5}=Z-1.37$

$\Rightarrow Z=20.15\approx 20$

$\therefore$The element is calcium. Hence, $\left(C\right)$ is the correct answer.

Why this question? Key Concept: This question gives an idea about the application of Moseley’s law.