Question

A certain element emits K_α X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.

(Use Planck constant h = 6.63 × 10^-34 Js= 4.14 × 10^-15 eVs, speed of light c = 3 × 10⁸ m/s.)

Answer 1

Given:
Energy of K_α X-ray, E = 3.69 keV = 3690 eV

Wavelength (λ) is given by

`λ = (hc)/E`

`λ = 1242/3690`

`λ = 0.33658  "nm"`

⇒ `λ = 0.34 xx 10^-9  "m"`

From Moseley's equation,

`sqrt(c/lambda) = a(Z - b)`

Here, c = speed of light
         `lambda` = wavelength of light
        Z = atomic number of element

On substituting the respective values,

`sqrt((3 xx 10^8)/(0.34 xx 10^-9))` = `5 xx 10^7(Z - 1.39)`

`⇒ sqrt(8.82 xx 10^17) = 5 xx 10^7(Z - 1.39)`

`⇒ 9.39 xx 10^8 = 5 xx 10^7(Z - 1.39)`

`⇒ 93.9/5 = Z - 1.39`

`⇒ Z = 93.9/5 + 1.39`

= 20.17 = 20

So, the element is calcium.