Question

The K_α X-rays of aluminium (Z = 13) and zinc (Z = 30) have wavelengths 887 pm and 146 pm respectively. Use Moseley's  law √v = a(Z − b) to find the wavelengths of the K_α X-ray of iron (Z = 26).

(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:
Wavelength of K_α X-rays of aluminium, λ₁ = 887 pm
Frequency of X-rays of aluminium is given by `nu_a = c/lambda`

`nu_a = (3 xx 10^8)/(887 xx 10^-12)`

`nu_a = 3.382 xx 10^17`

`nu_a = 33.82 xx 10^16  "Hz"`

Wavelength of K_α X-rays of zinc,  `lambda_2`= 146 pm
Frequency of X-rays of zinc is given by

`nu_z = (3 xx 10^8)/(146 xx 10^-12)`

`nu_z = 0.02055 xx 10^20`

`nu_z = 2.055 xx 10^18  "Hz"`

We know

`sqrt(nu) = a(Z-b)`

For aluminium,

`5.815 xx 10^8 = a(13 - b)  ...(1)`

For zinc,

`1.4331 xx 10^9 = a(30-b)  ...(2)`

Dividing (1) by (2)

`(13 - b)/(30 - b) = (5.815 xx 10^-1)/1.4331`

= 0.4057

`⇒ 30 xx 0.4057 - 0.4057  b = 13 - b`

`⇒ 12.171 - 0.4057  b + b = 13`

`b = 0.829/0.5943 = 1.39491`

`a = (5.815 xx 10^8)/11.33`

`= 0.51323 xx 10^8 = 5 xx 10^7`

For Fe,
Frequency  (`nu^'`) is given by

`nu^'` = 5× 10⁷ (26 − 1.39)
  = 5 × 24.61 × 10⁷
  = 123.05 × 10⁷

`nu^' = c/lambda`

Here, c =  speed of light
            `lambda` = Wavelength of light

`therefore  c/lambda = 5141.3 xx 10^14`

`⇒ lambda = (3 xx 10^8)/(5141.3 xx 10^14)`

`= 0.000198 xx 10^-5  "m"`

`= 198 xx 10^-12 = 198  "pm"`