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. Its orbital angular momentum is $3 h / 2 π$ . It is given that $h$ is Planck’s constant and $R$ is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are):

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$\frac{9}{5 R}$

$\frac{9}{32 R}$

$\frac{9}{3 R}$

$\frac{9}{16 R}$

answer is A, C.

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Detailed Solution

For the given orbit, we use
$4.5 a_{0} = a_{0} \frac{n^{2}}{Z}$
From Bohr’s second postulate, we use
$\frac{n h}{2 π} = \frac{3 h}{2 π}$
From above equations, we get
$n = 3 a n d Z = 2$
Thus possible wavelength are
$\frac{1}{λ_{1}} = R Z^{2} [\frac{1}{1^{2}} - \frac{1}{3^{2}}]$
$\Rightarrow λ_{1} = \frac{9}{32 R}$
and $\frac{1}{λ_{2}} = R Z^{2} [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$
$\Rightarrow λ_{2} = \frac{1}{3 R}$
$\frac{1}{λ_{3}} = R Z^{2} [\frac{1}{2^{2}} - \frac{1}{3^{2}}]$
$\Rightarrow λ_{3} = \frac{9}{5 R}$