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Standard XII
Physics
Davisson-Germer Experiment
Question

Electrons accelerated by potential V are diffracted from a crystal. If d= 1˚A and i−30o, V should be about (h=6.6×10−34Js,me=9.1×10−31kg,e=1.6×10−19C)
213590_bab0bea41851475d8e20d13c094b41bd.png
  1. 2000 V
  2. 50 V
  3. 1000 V
  4. 500 V

A
1000 V
B
2000 V
C
50 V
D
500 V
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Solution
Verified by Toppr

For constructive interference, 2dcosi=nλ and also wavelength, λ=h√2meV
So, 2dcosi=n×h√2meV
Squaring both sides, 4d2cos2i=n2h22meV

V=n2h28d2mecos2i=(1)2(6.6×10−34)28(1×10−10)2(9.1×10−31)(1.6×10−19)(cos230)=49.8V∼50V (we assume first order interference and take n=1)

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Similar Questions
Q1
Electrons accelerated by potential V are diffracted from a crystal. If d= 1˚A and i−30o, V should be about (h=6.6×10−34Js,me=9.1×10−31kg,e=1.6×10−19C)
213590_bab0bea41851475d8e20d13c094b41bd.png
View Solution
Q2
Electron accelerated by potential V are diffracted from a crystal. If d=1A and i=30∘. V should be about h=6.6×10−24Jsme=9.1×10−33kg.e=1.6×10−19C
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Q3
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 figure).
Electrons accelerated by potential V are diffracted from a crystal. If d=1Ao and i=30∘ , V should be about :
(h=6.6×10−34js,m=9.1×10−31kg,e=1.6×10−19c)
4605_429b816cb4b74f5e9b354e15ffa2062d.png
View Solution
Q4
De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: (|e|=1.6×10−19C,me=9.1×10−31kg,h=6.6×10−34Js).
View Solution
Q5
The de broglie wavelength of an electron (initially at rest) accelerated by an electric potential 'V' volts is given by:
Given: me=9.1×10−31 kg, e=1.6×10−19 C,
h=6.6×10−34 Js
View Solution