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Standard XII

Physics

Moseley's Law

The K α and K...

Question

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

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Solution

$K_{α} = E_{K} - E_{L}$ ...(1)

$λ K_{α} = 0.71 A^{\circ}$

$λ K_{β} = 0.63 A^{\circ}$

$K_{β} = E_{K} - E_{M}$ ...(2)

$L_{α} = E_{L} - E_{M}$ ...(3)

Substracting (2) from (1)

$K_{α} - K_{β} = E_{M} - E_{L} = - L_{α}$

or $L_{α} = K_{β} - K_{α}$

= $\frac{3 \times 10^{8}}{0.63 \times 10^{- 10}} - \frac{3 \times 10^{8}}{0.71 \times 10^{- 10}}$

= $4.761 \times 10^{18} - 4.225 \times 10^{18}$

= $0.526 \times 10^{18}$ Hz

Again $λ = \frac{3 \times 10^{8}}{0.536 \times 10^{18}}$

= $5.6 \times 10^{- 10} = 5.6 A^{\circ}$

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The Kα and Kβ X -rays of molybdenum have wavelengths 0.71 A∘ and 0.63 A∘ respectively. Find the wavelength of Lα X-ray of molybdenum.

Kα=EK−EL ...(1) λKα=0.71A∘ λKβ=0.63A∘ Kβ=EK−EM ...(2) Lα=EL−EM ...(3) Substracting (2) from (1) Kα−Kβ=EM−EL=−Lα or Lα=Kβ−Kα = 3×1080.63×10−10−3×1080.71×10−10 = 4.761×1018−4.225×1018 = 0.526×1018 Hz Again λ=3×1080.536×1018 = 5.6×10−10=5.6A∘

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