Question

What is the wavelength in picometers for an electron with 360 eV of kinetic energy?

Answer 1

Answer:

Wavelength = 64.635 pm

Explanation:

The expression for the deBroglie wavelength and kinetic energy is:

$\lambda=\frac {h}{\sqrt{2\times m\times K.E.}}$

Where,  

$\lambda$ is the deBroglie wavelength  

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

K.E. is the kinetic energy of the electron.

Given, K.E. = 360 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, K.E. = $360\times 1.6022\times 10^{-19}\ J=5.76792\times 10^{-17}\ J$

Applying in the equation as:

$\lambda=\frac {h}{\sqrt{2\times m\times K.E.}}$

$\lambda=\frac{6.626\times 10^{-34}}{\sqrt {{2\times 9.11\times 10^{-31}\times 5.76792\times 10^{-17}}}}\ m$

$\lambda=\frac{10^{-34}\times \:6.626}{\sqrt{10^{-48}\times \:105.0915024}}\ m$

$\lambda=6.4635\times 10^{-11}\ m$

Also, 1 m = 10¹² pm

So, Wavelength = 64.635 pm