Question

What is de-Broglie wavelength of electron having energy 10keV

Answer 1

De-Broglie wavelength is given by

$l=\frac{h}{mv}$, where $h$ is Planck's constant, $m$ mass of electron and $v$ speed of the electron.

Now,

$\frac{1}{2} mv^2=E\\ mv^2=2E\\ m^2v^2=2mE\\ mv=\sqrt{2mE}$

The De-Broglie wavelength is

$l=\frac{h}{\sqrt{2mE}} \\ l=\frac{6.63*10^{-34}}{\sqrt{2*9.11*10^{-31}*10^4*1.602*10^{-19}}} \\ l=1.23*10^{-11}$

The De-Broglie wavelength is $1.23*10^{-11} \;m$

Answer 2

Answer:

De-Borglie wavelength, $\lambda=1.22\times 10^{-11}\ m$

Explanation:

It is given that,

Energy of electron, E = 10 KeV = 10⁴ eV = 1.6 × 10⁻¹⁵ J

The relation between the wavelength and the energy of the electron is :

$\lambda=\dfrac{h}{\sqrt{2mE}}$

Where

h is the Planck's constant

m is the mass of electron

$\lambda=\dfrac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-15}} }$

$\lambda=1.22\times 10^{-11}\ m$

Hence, this is the required solution.