In Hydrogen spectrum the shortest wavelength in Balmer series is λ . The shortest wavelength in the Brackett series will be

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2λ

4λ

9λ

16λ

answer is B.

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Detailed Solution

For shortest wavelength in Balmer series, n1=2, n2=∞1λ=R142−1∞⇒λ=4RFor shortest wavelength in Brackett series, n1=4, n2=∞∴1λ'=R142−1∞⇒λ'=16R=4λ

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