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Which one of the following transitions of an electron in hydrogen atom emits radiation of the lowest wavelength ?

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$n_{2} = \infty to n_{1} = 2$

$n_{2} = 2 to n_{1} = 1$

$n_{2} = 5 to n_{1} = 3$

$n_{2} = 4 to n_{1} = 3$

answer is C.

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Detailed Solution

Emission of shortest wavelength means maximum energy

Maximum difference in the energy is between n = 2 to n = 1

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