The K_{α} and K_{β} X-rays of molybdenum have wavelengths 0.71 A and 0.63 A respectively. Find the wavelength of L_{α} X-ray of molybdenum.

(Use Planck constant h = 6.63 × 10^{-34} Js= 4.14 × 10^{-15} eVs, speed of light c = 3 × 10^{8} m/s.)

#### Solution

Given:-

Wavelength of K_{α}_{ }X-rays of molybdenum,`lambda_a = 0.71 "A"`

Wavelength of K_{β} X-rays of molybdenum,`lambda_b = 0.63 "A"`

Energy of K_{α}X-rays (`K_a`)is given by

`K_a = E_K - E_L` .....(1)

Energy of K_{β} X-rays (`K_β`) is given by

`K_β = E_K - E_L` .....(2)

Energy of L_{a} X-ray (`L_a`) is given by

`K_L = E_L - E_M`

Subtracting (2) from (1),

`K_α - K_β = E_M - E_L = -K_L`

or `K_L = K_β - K_α`

`K_L = (3 xx 10^8)/(0.63 xx 10^-10) - (3 xx 10^8)/(0.71 xx 10^-10)`

`K_L = 4.761 xx 10^-18 - 4.225 xx 10^18`

`K_L = 0.536 xx 10^18 "Hz"`

Again , `lambda = (3 xx 10^8)/(0.536 xx 10^-18)`

⇒ `lambda = 5.6 xx 10^-10 = 5.6 Å`