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=3 and n=2 states
B. n=3 and n=1 state
C. n=2 and n=1 state
D. n=4 and n=3 states

Answer 1

Correct Answer - D
We know that, the wavelength of emited radiation is givne by
`(1)/(lambda)R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))=RxxK"where,"K=(1)/(n_(1)^(2))-(1)/(n_(2)^(2))`
For `lambda_("max")` K should be minimum.
Which corresponds to ` n_(1)=3,n_(2)=4` So, transition takes place between n=4 and n=3 states.