1. 16λ 2. 2λ 3. 4λ 4. 9λ

Answer: 3. 4λ The shortest wavelength of a series comes from the transition from n=∞ to the series' lower level n₁ : dfrac1λ=R/n₁² , so λ∝ n₁² . Balmer: n₁=2 , so λ = 4/R . Brackett: n₁=4 , so λ' = 16/R = 4λ . The answer is 4λ .

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

In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Bracket series is :

1. \(16\lambda\)

2. \(2\lambda\)

3. \(4\lambda\)

4. \(9\lambda\)

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Answer: 3. \(4\lambda\)

The shortest wavelength of a series comes from the transition from \(n=\infty\) to the series' lower level \(n_1\): \(\dfrac1{\lambda}=\dfrac{R}{n_1^2}\), so \(\lambda\propto n_1^2\).

Balmer: \(n_1=2\), so \(\lambda = 4/R\).
Brackett: \(n_1=4\), so \(\lambda' = 16/R = 4\lambda\).

The answer is \(4\lambda\).

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In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Bracket series is :

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