#### 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.

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#### 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.

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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 courses CBSE (Arts), CBSE (Commerce), CBSE (Science)