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n = 2 to n = 1 transition of the following emit a photon with smallest wave length.

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He⁺

Li²⁺

Be³⁺

answer is D.

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Detailed Solution

According to Rydberg equation:

λ₁=R_Hz²(1/n₁²−1/n₂²)

As from the equation, the atomic number and wavelength are inversely related. More the atomic number, the shortest will be the wavelength.

The atomic number of Be³⁺is 4 and is the highest of all. So, this nucleus will have the shortest wavelength

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