A hydrogen atom makes a transition from $n=5$ to $n=1$ orbit. The wavelength of the photon emitted is $\lambda$. The wavelength of the photon emitted when it makes a transition from $n=5$ to $n=2$ orbit is

CBSE Class 12 · 2023 · Modern PhysicsMultiple Choice

A hydrogen atom makes a transition from

n = 5

n = 1

orbit. The wavelength of the photon emitted is

λ

. The wavelength of the photon emitted when it makes a transition from

n = 5

n = 2

orbit is

Answer: (D)

Energy of a transition:

Δ E = 13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) eV

, and

λ \propto \frac{1}{Δ E}

Δ E_{5 \to 1} = 13.6 (1 - \frac{1}{25}) = 13.6 \cdot \frac{24}{25}, Δ E_{5 \to 2} = 13.6 (\frac{1}{4} - \frac{1}{25}) = 13.6 \cdot \frac{21}{100} .

\frac{λ _{5 \to 2}}{λ _{5 \to 1}} = \frac{Δ E _{5 \to 1}}{Δ E _{5 \to 2}} = \frac{24/25}{21/100} = \frac{24 \times 4}{21} = \frac{32}{7} .

Hence

λ_{5 \to 2} = \frac{32}{7} λ

(D) \frac{32}{7} λ

CBSE Class 12 Board · 2023 · Single correct option