Multiple Choice

The radius of the orbit of an electron in a Hydrogen-like atom is $$4.5 $$$$\mathrm{a}_{0}$$ where $$\mathrm{a}_{0}$$ is the Bohr radius. Its orbital angular momentum is $$\displaystyle \frac{3\mathrm{h}}{2\pi}$$ It is given that $$\mathrm{h}$$ is Planck's constant and $$\mathrm{R}$$ is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)

A$$\displaystyle \frac{9}{32\mathrm{R}}$$
Correct Answer
B$$\displaystyle \frac{9}{16\mathrm{R}}$$
C$$\displaystyle \frac{9}{5\mathrm{R}}$$
Correct Answer
D$$\displaystyle \frac{4}{3\mathrm{R}}$$

Solution

Given data:
$$4.5\displaystyle \mathrm{a}_{0}=\mathrm{a}_{0}\frac{\mathrm{n}^{2}}{\mathrm{Z}}$$........... ($$\mathrm {i}$$)

$$\displaystyle \frac{\mathrm{n}\mathrm{h}}{2\pi}=\frac{3\mathrm{h}}{2\pi}$$ ..........(ii)

So, $$\mathrm{n}=3$$ and $$\mathrm{z}=2$$
So, possible wavelength are:
$$\displaystyle



\frac{1}{\lambda_{1}}=\mathrm{R}\mathrm{Z}^{2}[\frac{1}{1^{2}}-\frac{1}{3^{2}}]\Rightarrow \lambda_{1}=\frac{9}{32\mathrm{R}}$$

$$\displaystyle\frac{1}{\lambda_{2}}=\mathrm{R}\mathrm{Z}^{2}[\frac{1}{1^{2}}-\frac{1}{2^{2}}]\Rightarrow \lambda_{2}=\frac{1}{3\mathrm{R}}$$

$$\displaystyle\frac{1}{\lambda_{3}}=\mathrm{R}\mathrm{Z}^{2}[\frac{1}{2^{2}}-\frac{1}{3^{2}}]\Rightarrow \lambda_{3}=\frac{9}{5\mathrm{R}}$$

SIMILAR QUESTIONS

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