The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a 0 , where a 0 is the Bohr radius.…

1. $\frac{9}{32\text{R}}$
2. $\frac{9}{16\text{R}}$
3. $\frac{9}{5\text{R}}$
4. $\frac{4}{3\text{R}}$

Solution

${\text{R}}_{\text{n}}=\text{4.5}{\text{a}}_0 \Rightarrow {\text{a}}_0{\text{n}}^2/\text{z}=\text{4.5 }{\text{a}}_0$
$\therefore \text{n}=3, \text{z}=2$
$\text{L}=\text{mvr}=\frac{3\text{h}}{2\pi }\left[\text{as n}=3\text{,}\text{z}=2\right]$

$\frac{1}{\lambda }={\text{Rz}}^2\left(\frac{1}{{\text{n}}_{\text{f}}^2}-\frac{1}{{\text{n}}_{\text{1}}^2}\right)$
$\frac{1}{{\lambda }_{3\to 1}}=\text{R}4\left[\frac{1}{1}-\frac{1}{9}\right]=4\text{R}\frac{8}{9}\Rightarrow {\lambda }_{3\to 1}=\frac{9}{32\text{R}}$
$\frac{1}{{\lambda }_{2\to 1}}=\text{R}4\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3}{4}4\text{R}\Rightarrow {\lambda }_{2\to 1}=\frac{1}{3\text{R}}$
$\frac{1}{{\lambda }_{3\to 2}}=\text{R}4\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5}{36}4\text{R}\Rightarrow {\lambda }_{3\to 2}=\frac{9}{5\text{R}}$

Asked in: JEE Advanced 2013 (Paper 2)

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