The `K`-absorption edge of an unknown element is `0.171 Å`
a. Identify the element.
b. Find the average wavelength of the `K_(alpha),K_(alpha),K_(beta),K_(gamma)` lines.
c.
If a `100 e V` electron strikes the target of this element , what is the minimum wavelength of the x-ray emitted?

Explanation

From Moseley's law , the wavelength of `K` series of X-rays is given by taking modified Rydberg's given as
`(1)/(lambda) = R (Z - 1)^(2) (1 - (1)/(n^(2)))` for `K` lines, where `n = 2,3,4` ,…..
a. For `K` -absorption edge, we put `n = oo` in above expression to get
`(Z - 1) - sqrt((1)/(lambda R))`
or `Z = sqrt((1)/((0.171 xx 10^(-10)) (1.097 xx 10^(7)))) + 1 = 74`
element is tungsten.
b. For `K_(alpha) line: (1)/(lambda _(K_alpha)) = R(74 - 1)^(2) [1 - (1)/(2^(2))]`
`implies lambda _(K_alpha) = 0.228 Å`
For `K_(beta) "line"`: (1)/(lambda _(K_beta)) = R(74 - 1)^(2) [1 - (1)/(3^(2))]`
`implies lambda _(K_beta) = 0.192 Å`
For `K_(gamma) line: (1)/(lambda _(K_gamma)) = R(74 - 1)^(2) [1 - (1)/(4^(2))]`
`implies lambda _(K_gamma) = 0.182 Å`
c. The shortest wavelength corresponding to an electron with kinetic energy `100 e V` is given by
`lambda _(c) = (hc)/(E) = (12431)/(100) Å = 124.31 Å`