Modern Physics Ques 150
- 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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Answer:
Correct Answer: 150.(b)
Solution:
Formula:
- $K _{\alpha}$ 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, $b=1$
$\therefore \quad \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}} $
$=\quad \frac{41 \times 41}{28 \times 28}=\frac{1681}{784}=2.144$
BETA
