The first order reflection n-1 from a crystal of the X-ray from a copper anode tube 1-54 A occurs at an angle of 45- What is the distance between the set of plane causing the diffraction-

Given: n=1 λ=1.54 A ϕ=45^∘ From Bragg's Law, 2dsinϕ=nλ ⇒ 2× d×sin45^∘=1×1.54 nbsp; ⇒ d=1.1 A Hence, option (A) is the correct answer. ...

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Standard XII

Chemistry

Bragg's Equation

The first ord...

Question

The first order reflection $(n = 1)$ from a crystal of the X-ray from a copper anode tube $(1.54 ˚ A)$ occurs at an angle of $45^{\circ}$ . What is the distance between the set of plane causing the diffraction?

1.1 ˚ A

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1.3 ˚ A

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1.5 ˚ A

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1.7 ˚ A

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Solution

The correct option is A $1.1 ˚ A$
Given:
$n = 1$
$λ = 1.54 ˚ A$
$ϕ = 45^{\circ}$

From Bragg's Law,

$2 d sin ϕ = n λ$

$\Rightarrow 2 \times d \times sin 45^{\circ} = 1 \times 1.54$

$\Rightarrow d = 1.1 ˚ A$

Hence, option $(A)$ is the correct answer.

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The first order reflection (n=1) from a crystal of the X-ray from a copper anode tube (1.54 ˚A) occurs at an angle of 45∘. What is the distance between the set of plane causing the diffraction?

The correct option is A 1.1 ˚AGiven: n=1 λ=1.54 ˚A ϕ=45∘ From Bragg's Law, 2dsinϕ=nλ ⇒2×d×sin45∘=1×1.54 ​​​​​​​⇒d=1.1 ˚A ​​​​​​​Hence, option (A) is the correct answer.

The first order reflection $(n = 1)$ from a crystal of the X-ray from a copper anode tube $(1.54 ˚ A)$ occurs at an angle of $45^{\circ}$ . What is the distance between the set of plane causing the diffraction?

The correct option is A $1.1 ˚ A$
Given:
$n = 1$
$λ = 1.54 ˚ A$
$ϕ = 45^{\circ}$

From Bragg's Law,

$2 d sin ϕ = n λ$

$\Rightarrow 2 \times d \times sin 45^{\circ} = 1 \times 1.54$

$\Rightarrow d = 1.1 ˚ A$

Hence, option $(A)$ is the correct answer.