To obtain a diffraction peak, for a crystalline solid with interplane distance equal to the wavelength of incident X-ray radiation, the angle of incidence should be:

90^{\circ}

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0^{\circ}

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30^{\circ}

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60^{\circ}

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Solution

The correct option is D $30^{\circ}$
Accrording to Bragg's equation

$n λ = 2 d s i n θ$

Given $λ = d$

$n = 2 s i n θ$

$1 = 2 s i n θ$ , for n=1

$θ = 30^{o}$

Option C is correct.

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Similar questions

Q. The distance between planes of ions parallel to the face of the unit cell of sodium chloride is

282 p m

. The smallest angle of reflection observed in the X-ray diffraction pattern is

{4.97}^{\circ}

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Calculate the distance between the planes of ions if the smallest angle of reflection is x-ray diffraction is ${5.97}^{0}$ and the wavelength of incident ray is 58.6 pm? $(s i n {5.97}^{0}$ = 0.104)

The angle between an incident ray and the plane mirror is $30^{\circ}$ . The total angle between the incident ray and reflected ray will be : (a) $30^{\circ}$ (b) $60^{\circ}$ (c) $90^{\circ}$ (d) $120^{\circ}$

Q. If a strong diffraction peak is observed when electron are incident at an angle. 'i' from the normal to the crystal planes with distance 'd' wavelength them (see figure) de Broglie wavelength

λ_{d B}

of electrons can be calculate by the relationship (n is an integer)

Q. An X-ray beam of wavelength

1.0 ˚ A

is incident on a crystal of lattice spacing

2.8 ˚ A

. The value of Bragg angle for first order of diffraction is :

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To obtain a diffraction peak, for a crystalline solid with interplane distance equal to the wavelength of incident X-ray radiation, the angle of incidence should be:

The correct option is D 30∘Accrording to Bragg's equationnλ=2d sin θGiven λ=dn=2 sin θ1=2 sin θ, for n=1θ=30oOption C is correct.

The correct option is D $30^{\circ}$
Accrording to Bragg's equation

$n λ = 2 d s i n θ$

Given $λ = d$

$n = 2 s i n θ$

$1 = 2 s i n θ$ , for n=1

$θ = 30^{o}$

Option C is correct.