The largest distance between the inter atomic planes of a crystal is $10^{- 7} c m$ . The upper limit for the wavelength of X-rays which can be usefully studied with this crystal is

1 ˙ A

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2 ˙ A

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10 ˙ A

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20 ˙ A

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Solution

The correct option is D $20 ˙ A$
Bragg's law, $2 d s i n θ = n λ$ or $λ = \frac{2 d s i n θ}{n}$
For maximum wavelength, $n_{m i n} = 1, (s i n θ)_{m a x} = 1$
$∴ λ_{m a x} = 2 d$ or $λ_{m a x} = 2 \times 10^{- 7} c m = 20 ˙ A$

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The largest distance between the inter atomic planes of a crystal is 10−7cm. The upper limit for the wavelength of X-rays which can be usefully studied with this crystal is

The correct option is D 20˙ABragg's law, 2dsinθ=nλ or λ=2dsinθn For maximum wavelength, nmin=1,(sinθ)max=1 ∴λmax=2d or λmax=2×10−7 cm=20˙A

The largest distance between the inter atomic planes of a crystal is $10^{- 7} c m$ . The upper limit for the wavelength of X-rays which can be usefully studied with this crystal is

The correct option is D $20 ˙ A$
Bragg's law, $2 d s i n θ = n λ$ or $λ = \frac{2 d s i n θ}{n}$
For maximum wavelength, $n_{m i n} = 1, (s i n θ)_{m a x} = 1$
$∴ λ_{m a x} = 2 d$ or $λ_{m a x} = 2 \times 10^{- 7} c m = 20 ˙ A$