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Question: When an electron in an excited Mo atom falls from L to the K shell, an X-ray is emitted. These X-ray...

When an electron in an excited Mo atom falls from L to the K shell, an X-ray is emitted. These X-rays are diffracted at angle of 7.75o7.75^{o} by planes with a separation of 2.64Å. What is the difference in energy between K-shell and L-shell in Mo assuming a first-order diffraction. (sin⁡7.75o=0.1349)(\sin 7.75^{o} = 0.1349)

A

36.88 ×10−15J\times 10^{- 15}J

B

27.88×10−16J27.88 \times 10^{- 16}J

C

63.88 ×10−17J\times 10^{- 17}J

D

64.88×10−16J64.88 \times 10^{- 16}J

Answer

27.88×10−16J27.88 \times 10^{- 16}J

Explanation

Solution

2dsin⁡θ=nλ2d\sin\theta = n\lambda

λ=2dsin⁡θ=2×2.64×10−10×sin⁡7.75o=0.7123×10−10mE=hcλ=6.62×10−34×3×1080.7123×10−10=27.88×10−16J\lambda = 2d\sin\theta = 2 \times 2.64 \times 10^{- 10} \times \sin 7.75^{o} = 0.7123 \times 10^{- 10}mE = \frac{hc}{\lambda} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{0.7123 \times 10^{- 10}} = 27.88 \times 10^{- 16}J