1. 9/16λ 2. 20/7λ 3. 20/13λ 4. 16/25λ

Answer: 2. 20/7λ Use dfrac1λ=R(dfrac1n₁²-dfrac1n₂²) . 3 to 2: dfrac1λ=R(1/4-1/9)=5R/36 . 4 to 3: dfrac1λ'=R(1/9-dfrac116)=7R/144 . λ'/λ=5/36/7/144=5×144/36×7=20/7 So λ'=20/7λ . Option (2).

Topic: Bohr model, energy levels, hydrogen spectrum· Atoms and Nuclei

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

1. $\frac{9}{16}\lambda$

2. $\frac{20}{7}\lambda$

3. $\frac{20}{13}\lambda$

4. $\frac{16}{25}\lambda$

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Answer: 2. $\frac{20}{7}\lambda$

Use $\dfrac1\lambda=R\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right)$.

3 to 2: $\dfrac1\lambda=R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{36}$.
4 to 3: $\dfrac1{\lambda'}=R\left(\dfrac19-\dfrac1{16}\right)=\dfrac{7R}{144}$.

$$\frac{\lambda'}{\lambda}=\frac{5/36}{7/144}=\frac{5\times144}{36\times7}=\frac{20}{7}$$

So $\lambda'=\tfrac{20}{7}\lambda$. Option (2).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

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