Electrons accelerated by potential V are diffracted from a crystal. If \(\mathrm{d}=1 \AA\) and \(\mathrm{i}=30^{\circ}, \mathrm{V}\) should be about (\(\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\))

JEE Main · 2008 · 27 Apr 2008 · Offline · Modern Physics

Passage

Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see in figure).

Electrons accelerated by potential V are diffracted from a crystal. If

d = 1 \overset{˚}{A}

and

i = 3 0^{\circ}, V

should be about (

h = 6.6 \times 1 0^{- 34} Js, m_{e} = 9.1 \times 1 0^{- 31} kg, e = 1.6 \times 1 0^{- 19} C

)

(A)2000 V
(B)50 V✓ Correct
(C)500 V
(D)1000 V

Answer: (B)

Solution

The diffraction condition for the angle measured from the normal is

2 d cos i = nλ

, and the de Broglie wavelength after acceleration through a potential

V

λ = \frac{h}{2 m e V}

. Taking the first order and combining,

2 d cos i = \frac{h}{2 m e V}

, which solves to

V = \frac{h ^{2}}{8 m e d ^{2} c o s ^{2} i}

. Substituting

d = 1 0^{- 10} m

i = 3 0^{\circ}

so that

cos^{2} i = \frac{3}{4}

, together with the given constants, the accelerating voltage comes out to about

50 V

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option

Practise more: all JEE Main PYQs on Modern Physics →

JEE Main · 2008 · 27 Apr 2008 · Offline · Modern Physics

Solution

More JEE Main questions on Modern Physics