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Solution :

The frequency of radiation emitted is given by <br> `v = ©/(lambda) = K((1)/(n_(1)^(2))- (1)/(n_(2)^(2)))`<br> Thus `2.467 xx 10^(15) Hz= K((1)/(1^(2)) - (1)/(2^(2)))`<br> or` K = (4)/(3) xx 2.467 xx 10^(15) Hz`<br> The frequency of the radiation emitted in the transition <br> `n = 3 to n= 3` is <br> `v^(1) = K[(1)/(1^(2)) - (1)/(3^(2))]`<br>` = (8)/(9)K= (8)/(9)2.467 xx 10^(15) Hz` <br>` = 2.92 xx 10^(15)Hz`.