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 betweenA. n=3 and n=2 statesB. n=3 and n=1 stateC. n=2 and n=1 stateD. n=4 and n=3 states

The ionization enegry of the electron in the hydrogen atom in its grou

1.	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 betweenA. n=3 and n=2 statesB. n=3 and n=1 stateC. n=2 and n=1 stateD. n=4 and n=3 states
Answer» 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.