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