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The ionization enegry of the electron in...

The ionization enegry of the electron in the hydrogen atom in its ground state is `13.6 ev`. The atoms are excited to higher energy levels to emit radiations of `6` wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between

A

n=3 to n=1 states

B

n=4 to n=3 states

C

n=3 to n=2 states

D

n=2 to n=1 states

Text Solution

Verified by Experts

The correct Answer is:
B

Number of wavelelngth =`(n(n-b))/(2)`
where, n=number of orbit from which transition place
`:. " " 6=(n(n-1))/(2)rArrn=4`
`:.` The wavelength of emitted radiations will be maximum for transition n=4 to n=3
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