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The wavelength of the first member of the balmer series of hydrogen is `6563xx10^(-10)m`. Calculate the wavelength of its second member. |
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Answer» `(1)/(lamda_(1))=R_(H)[(1)/(2^(2))-(1)/(3^(2))] and (1)/(lamda_(2))=R_(H)[(1)/(2^(2))-(1)/(4^(2))]` `(lamda_(2))/(lamda_(1))=(5)/(36)xx(16)/(3)=(20)/(27)` `lamda_(2)=(20)/(27)xx6563xx10^(-10)=4861xx10^(-10)m`. |
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