Question

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 = 4 to n = 3 states
B. n = 3 to n = 2 states
C. n = 3 to n = 1 states
D. n = 2 to n = 1 states

Answer 1

Correct Answer - A
`(n(n-1))/(2)=6 `
`implies n=4 `
For maximum wavelength energy difference betweeen states should be minimum because
`lambda=(hc)/(DeltaE)`
So , transition state in `n=4` to n=3