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Question: The de Broglie wavelength of an electron in a metal 27<sup>0</sup>C is : (Given \(m_{e} = 9.1 \time...

The de Broglie wavelength of an electron in a metal 270C is :

(Given me=9.1×10−31kg,kB=1.38×10−23Jk−1m_{e} = 9.1 \times 10^{- 31}kg,k_{B} = 1.38 \times 10^{- 23}Jk^{- 1})

A

6.2×10−9m6.2 \times 10^{- 9}m

B

6.2×10−10m6.2 \times 10^{- 10}m

C

6.2×10−8m6.2 \times 10^{- 8}m

D

6.2×10−7m6.2 \times 10^{- 7}m

Answer

6.2×10−9m6.2 \times 10^{- 9}m

Explanation

Solution

: Here (;gk) T=27+273=300KT = 27 + 273 = 300K

For an electron in a metal momentum

p=3mkBTp = \sqrt{3mk_{B}T}

De Broglie wavelength of an electron is

λ=hp=h3mkBT\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mk_{B}T}}

=h3×(9.1×10−31)×(1.38×10−23)×300=6.2×10−9m= \frac{h}{\sqrt{3 \times (9.1 \times 10^{- 31}) \times (1.38 \times 10^{- 23}) \times 300}} = 6.2 \times 10^{- 9}m