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 time of the Bahmer series is given by ` overline(v) =(1)/(lambda)=R((1)/(2^2)-(1)/(n^2))=R//4`.
`therefore R =4//lambda =(4//3636)xx10^(10)m^(-1)`.
Further the wavelength of the `K_(a)` series are given by the relation.
`overlin(v) =(1)/(lambda)=R(Z-1)^(2)((1)/(1^2)-(1)/(n^2))`.
The maximum wave number correspondence to `n =prop` and , therefore, we must have `overline(v) =(1)/(lambda)= R(Z-1)^(2)`
or `(Z-1)^(2)=(1)/(R lambda)=(3646xx10^(-10))/(4xx1xx10^(-10))=911.5`
`therefore (Z-1)=sqrt(911.5)cong 30.2 ` or `Z=30.2cong 31`.
Thus the atomic number of the element concerned is 31.
The element having atomic number `Z=31` is Gallium.