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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 to n=1 statesB. n=4 to n=3 statesC. n=3 to n=2 statesD. n=2 to n=1 states |
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Answer» Correct Answer - 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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