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What is the wavelength of the radiation emitted to the electron in a hydrogen atom junps from `n = 1 to n = 2`?

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Verified by Experts

The energy of `n = 2` state is
`E_(2)= (-13.6eV)/(4)= -3.4eV`
The energy of `n = 00` state is zero.The wavelength emitted in the given transition is
`lambda = (hc)/(DeltaE) `
`= (1242eV nm)/(3.4eV) = 365nm`
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