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The wavelength of high energy transition of H atom is `91.2 nm` Calculate the corresponding wavelength of He atom |
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Answer» Correct Answer - B ` (1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)` For hydrogen atom `(1)/(91.2) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)` For `He^(Theta)` ` (1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)` Dividing the equation we get `(lambda)/(91.2) = (1)/(4)` `lambda = 22.8 nm` |
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