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Solution for Compute the Typical De Broglie Wavelength of an Electron in a Metal at 27 ºC and Compare It with the Mean Separation Between Two Electrons in a Metal Which is Given to Be About 2 × 10−10 M. - CBSE (Science) Class 12 - Physics

Question

Compute the typical de Broglie wavelength of an electron in a metal at 27 ºC and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10−10 m.

Solution

Temperature, T = 27°C = 27 + 273 = 300 K

Mean separation between two electrons, r = 2 × 10−10 m

De Broglie wavelength of an electron is given as:

`lambda = h/sqrt(3mkT)`

Where,

h = Planck’s constant = 6.6 × 10−34 Js

m = Mass of an electron = 9.11 × 10−31 kg

k = Boltzmann constant = 1.38 × 10−23 J mol−1 K−1

`:. lambda = (6.6 xx 10^(-34))/sqrt(3xx9.11 xx10^(-31) xx 1.38 xx 10^(-23) xx 300)`

`~~ 6.2 xx 10^(-9) m`

Hence, the de Broglie wavelength is much greater than the given inter-electron separation.

  Is there an error in this question or solution?

APPEARS IN

 NCERT Physics Textbook for Class 12 Part 2 (with solutions)
Chapter 11: Dual Nature of Radiation and Matter
Q: 36 | Page no. 411
Solution for question: Compute the Typical De Broglie Wavelength of an Electron in a Metal at 27 ºC and Compare It with the Mean Separation Between Two Electrons in a Metal Which is Given to Be About 2 × 10−10 M. concept: Wave Nature of Matter. For the course CBSE (Science)
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