First order X-ray reflection -154 pm is maximum from a set of 2-0-0 planes of a B-C-C lattice was observed at 16- 6 - Then- the edge length of unit cell isA- 360 pmB- 420 pmC- 560 pmD- 180 pm

The correct option is C 560 pm For, first order reflection, n = 1 λ =2d sin θ, d=λ/2 sin θ=154/2 × 0.2773=280pm d(2, 0, 0)=154/√(2^2+0^2+0^2)⇒ a=2 × 280 = 56 ...

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Standard XII

Chemistry

Bragg's Equation

First order X...

Question

First order X-ray reflection ( $λ$ = 154 pm) is maximum from a set of (2, 0, 0) planes of a B.C.C lattice was observed at $16^{\circ}$ , 6’. Then, the edge length of unit cell is

360 pm

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420 pm

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560 pm

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180 pm

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Solution

The correct option is C
560 pm

For, first order reflection, n = 1
$λ = 2 d s i n θ, d = \frac{λ}{2 s i n θ} = \frac{154}{2 \times 0.2773} = 280 p m d (2, 0, 0) = \frac{154}{\sqrt{2^{2} + 0^{2} + 0^{2}}} \Rightarrow a = 2 \times 280 = 560 p m$

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First order X-ray reflection (λ = 154 pm) is maximum from a set of (2, 0, 0) planes of a B.C.C lattice was observed at 16∘, 6’. Then, the edge length of unit cell is

The correct option is C 560 pm For, first order reflection, n = 1 λ=2d sinθ, d=λ2 sin θ=1542×0.2773=280pmd(2, 0, 0)=154√22+02+02⇒a=2×280=560 pm

First order X-ray reflection ( $λ$ = 154 pm) is maximum from a set of (2, 0, 0) planes of a B.C.C lattice was observed at $16^{\circ}$ , 6’. Then, the edge length of unit cell is

The correct option is C
560 pm

For, first order reflection, n = 1
$λ = 2 d s i n θ, d = \frac{λ}{2 s i n θ} = \frac{154}{2 \times 0.2773} = 280 p m d (2, 0, 0) = \frac{154}{\sqrt{2^{2} + 0^{2} + 0^{2}}} \Rightarrow a = 2 \times 280 = 560 p m$