InterviewSolution
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InterviewSolution
| 1. |
If the shorts series limit of the balmer series for hydrogen is `3644 Å`, find the atomic number of the element which gives X-ray wavelength down to `1 Å`. Identify the element. |
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Answer» The short series limit of the Balmar series is corresponding to transition `n = oo to n = 1` which is given by `(1)/(lambda) = R((1)/(2^(2)) - (1)/(oo^(2))) = ( R)/4` or `R = (4)/(lambda) = (4)/(3644) (Å) ^(-1)` The short wavelength corresponds to transition from `n = oo to n = 1` which is given as `(1)/(lambda) = R (Z- 1)^(2) [(1)/(1^(2)) - (1)/(oo^(2))]` or `(Z- 1)^(2) = (1)/(lambda_( c) R) = (1)/(1 Å xx (4)/(3644) (Å)^(-1)) = (3644)/(4) = 911` or `Z - 1 = 30.2` or `Z = 31.2 = 31`. Thus, the atomic number of the element is `31` which is gallium. |
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