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.

Explanation

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.