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The wavelength of the radiations emitted when in a hydrogen atom electron falls from infinity to stationary state is ` : (R_H = 1. 097 xx10^7 m^(-1))` .A. `406 nm`B. `192 nm`C. `91 nm`D. `9.1 xx 10^-8 nm` |
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Answer» Correct Answer - C ( c) `(1)/(lamda) = R[(1)/(n_1^2) - (1)/(n_2^2)]` `(1)/(lamda) = 1.097 xx 10^7 m^-1 [(1)/(1^2) - (1)/(oo^2)]` `:. lamda = 91 xx 10^-9 m` We know `10^-9 = 1nm` So `lamda = 91 nm`. |
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