Answer:
The wavelength of the electron is [tex]\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]
Explanation:
Given that,
Temperature = 300 K
We know that,
The energy of free electron is
[tex]E=\dfrac{(\hbar)^2k^2}{2m}[/tex]
[tex]k=\dfrac{\sqrt{2mE}}{\hbar}[/tex]
Where, k = wave number
The momentum of the electron is
[tex]p=\hbar k[/tex]
Th effective mass is
[tex]m=am_{0}[/tex]
We need to calculate the wavelength of the electron
Using formula of wave number
[tex]k=\dfrac{2\pi}{\lambda}[/tex]
[tex]\lambda=\dfrac{2\pi}{k}[/tex]
Put the value of k
[tex]\lambda=\dfrac{2\pi}{\dfrac{\sqrt{2mE}}{\hbar}}[/tex]
[tex]\lambda=\dfrac{h}{\sqrt{2am_{0}E}}[/tex]
We know that,
Thermal energy of electron
[tex]E=3kT[/tex]
The de Broglie wavelength of the electron is
[tex]\lambda=\dfrac{h}{\sqrt{2am_{0}\times3kT}}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times3\times300\times1.380\times10^{-23}\times a}}[/tex]
[tex]\lambda=\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]
Hence, The wavelength of the electron is [tex]\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]
