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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