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Q. No.For a crystal with \(\lambda = 1~\mathring{A}\) and Bragg's angle \(\theta= 60^{\circ}\), the second-order diffraction, \(d\) (distance between two consecutive atomic layers) will be:
1. \(1.15~\mathring{A}\)
2. \(0.75~\mathring{A}\)
3. \(0.55~\mathring{A}\)
4. \(2.1~\mathring{A}\)
1. \(1.15~\mathring{A}\)
2. \(0.75~\mathring{A}\)
3. \(0.55~\mathring{A}\)
4. \(2.1~\mathring{A}\)
Subtopic: Â Diffraction |
 53%
From NCERT
AIPMT - 1998
Hints
The interplanar distance in a crystal is \(2.8 \times 10^{-8}~\text{m}\). The value of the maximum wavelength which can be diffracted:
1. \(2.8 \times 10^{-8}~\text{m}\)
2. \(5.6 \times 10^{-8}~\text{m}\)
3. \(1.4 \times 10^{-8}~\text{m}\)
4. \(7.6 \times 10^{-8}~\text{m} \)
1. \(2.8 \times 10^{-8}~\text{m}\)
2. \(5.6 \times 10^{-8}~\text{m}\)
3. \(1.4 \times 10^{-8}~\text{m}\)
4. \(7.6 \times 10^{-8}~\text{m} \)
Subtopic: Â Diffraction |
From NCERT
AIPMT - 2001
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- 1(current)
Q. No.
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