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An electron in hydrogen atom experiences a transition from n = 6 to n = 3. Find out the frequency of the radiation. |
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Answer» Solution :WAVE number of radiation emitted in H atom, `(1)/(lambda)=R[(1)/(3^(2))-(1)/(6^(2))]=R[(1)/(9)-(1)/(36)]` `(1)/(lambda)=(27R)/(324)=(R)/(12)....(1)` `rArr` Now `c=lambda v` `:.(1)/(lambda)=(v)/(c)` `:.(v)/(c)=(R)/(12)[ :.` From EQUATION (1)] `:.v=(Rc)/(12)=(1.09xx10^(7)xx3xx10^(8))/(12)` `:.v=0.275xx10^(15)Hz` |
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