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.

Answer

`Z = 31` Gallium

Explanation

The short limit of Balmar series is given by ltbgt ` bar v = 1//lambda = R((1)/(2^(2)) - (1)/(oo^(2))) = R//4`
` :. R = 4//lambda = (4)/(3646) xx 10 ^(10) m^(-1)`
Further, the wavelength of the `K_(alpha)` series are given by the ralation `bar v = (1)/(lambda) = R (Z - 1)^(2) ((1)/(1^(2)) - (1)/(n^(2)))`
The maximum wave number corresponding to `n = oo` and therefore we must have
`bar v = (1)/(lambda) = R (Z - 1)^(2)`
or `Z - 1^(2) = (1)/(R lambda) = (3464 xx 10^(-19))/(4 xx 1 xx 10^(-10)) = 911.5`
`:. Z = 31.2 ~= 31`
Thus , the atomic number of the element concern is `31`. The element having atomic number `Z = 31` is gallium.