Question 605: Atoms & Nuclei

Question

The interplanar distance in a crystal is 2.8 $ 	imes $ 10
$-$8
m. The value of maximum wavelength which can be diffracted

Accepted Answer

2dsin$	heta $ = n$\lambda $
$ \because $ $ - 1 \le \sin 	heta  \le 1$
Therefore $\lambda $
max
= 2d
$ \Rightarrow $ $\lambda $
max
= 2 $ 	imes $ 2.8 $ 	imes $ 10
$-$8
= 5.6 $ 	imes $ 10
$-$8
m

Answer

2.8 $ 	imes $ 10
$-$8
m

Answer

5.6 $ 	imes $ 10
$-$8
m

Answer

1.4 $ 	imes $ 10
$-$8
m

Answer

7.6 $ 	imes $ 10
$-$8
m

Atoms & Nuclei

The interplanar distance in a crystal is 2.8 \times 10 -8 m. The value of maximum wavelength which can be diffracted

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