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The wavelength of the first line in blamer series in the hydrogen spectrum is `lambda`. What is the wavelength of the second line:A. `(20lambda)/27`B. `(3lambda)/16`C. `(5lambda)/36`D. `(3lambda)/4` |
Answer» Correct Answer - A `1/lambda_(1)= R(1/4-1/9) Rightarrow lambda_(1)=(4xx9)/(5R)` Similarly `1/lambda_(2)=R(1/4-1/4^(2))` `Rightarrowlambda_(2)=16/(3R)=16/3xx(5lambda)/(4xx9)=20/(27lambda)` |
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