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 $\frac{3 h}{2 \pi}$. It is given that $h$ is Planck constant and $R$ is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)

(A) $\frac{9}{32 R}$; (C) $\frac{9}{5 R}$

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Modern Physics

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

\frac{3 h}{2 π}

. It is given that

h

is Planck constant and

R

is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)

Answer: (A), (C)

From angular momentum:

\frac{nh}{2 π} = \frac{3 h}{2 π}

, so

n = 3

.

From radius:

r = a_{0} \frac{n ^{2}}{Z} = 4.5 a_{0}

\frac{9}{Z} = 4.5

, so

Z = 2

.

The atom is

He^{+}

with electron in

n = 3

. Possible transitions:

3→1:

\frac{1}{λ _{1}} = R Z^{2} [\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}}] = 4 R \cdot \frac{8}{9}

, so

λ_{1} = \frac{9}{32 R}

✓

3→2:

\frac{1}{λ _{2}} = 4 R [\frac{1}{4} - \frac{1}{9}] = 4 R \cdot \frac{5}{36}

, so

λ_{2} = \frac{9}{5 R}

✓

2→1:

\frac{1}{λ _{3}} = 4 R \cdot \frac{3}{4}

, so

λ_{3} = \frac{1}{3 R}

(not listed)

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options