Modern Physics 4 Question 10
10. If $\lambda _{Cu}$ is the wavelength of $K _{\alpha}, X$-ray line of copper (atomic number 29) and $\lambda _{MO}$ is the wavelength of the $K _{\alpha}$, X-ray line of molybdenum (atomic number 42 ), then the ratio $\lambda _{Cu} / \lambda _{Mo}$ is close to
(2014 Adv.)
(a) 1.99
(b) 2.14
(c) 0.50
(d) 0.48
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Solution:
- $K _{\alpha}$ transition takes place from $n _1=2$ to $n _2=1$
$$ \therefore \quad \frac{1}{\lambda}=R(Z-b)^{2} \frac{1}{(1)^{2}}-\frac{1}{(2)^{2}} $$
For $K$-series, $b=1$
$$ \begin{aligned} \therefore & \frac{1}{\lambda} \propto(Z-1)^{2} \\ \Rightarrow \quad \frac{\lambda _{Cu}}{\lambda _{Mo}} & =\frac{\left(z _{Mo}-1\right)^{2}}{\left(z _{Cu}-1\right)^{2}}=\frac{(42-1)^{2}}{(29-1)^{2}} \\ & =\frac{41 \times 41}{28 \times 28}=\frac{1681}{784}=2.144 \end{aligned} $$
