In a hydrogen atom modeled according to Bohr's postulates, an electron undergoes a transit… — JAMB UTME

In a hydrogen atom modeled according to Bohr's postulates, an electron undergoes a transition from an initial excited state $n$ to the ground state $n=1$ , emitting a photon whose wavelength is $\lambda = \frac{16}{15R}$ , where $R$ is the Rydberg constant. What is the orbital angular momentum of the electron in its initial excited state prior to the transition?

A
$\frac{h}{2\pi}$
B
$\frac{h}{\pi}$
$\frac{2h}{\pi}$ Answer
D
$\frac{4h}{\pi}$

Answer

\frac{2h}{\pi}

Using the Rydberg formula

\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)

with

\lambda = \frac{16}{15R}

, we find

\frac{15}{16} = 1 - \frac{1}{n^2}

, which gives

n = 4

. By Bohr's angular momentum quantization condition,

L = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}

Step-by-Step Solution

Apply the Rydberg formula for hydrogen to determine the principal quantum number

n

of the initial state.

\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \implies \frac{15R}{16} = R \left( 1 - \frac{1}{n^2} \right) \implies 1 - \frac{1}{n^2} = \frac{15}{16} \implies \frac{1}{n^2} = \frac{1}{16} \implies n = 4

The Rydberg formula links the photon wavelength emitted during a transition between energy levels to the principal quantum numbers.

Calculate the orbital angular momentum

L

for

n=4

using Bohr's quantization condition.

L = \frac{n h}{2\pi} = \frac{4 h}{2\pi} = \frac{2 h}{\pi}

Bohr's quantization postulate dictates that orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

Key Concept

Bohr's Atomic Model: Spectral transitions and angular momentum quantization

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