# Calculating de Broglie Wavelength for Proton

## What is the de Broglie wavelength for a proton with energy 50 keV?

Due to the limitations of Canvas, please give the wavelength in angstroms (x 10-10 m) to 2 sig figs.

## Answer:

1.2826 x 10^-13 m

The de Broglie wavelength for a proton with energy 50 keV is calculated by using the formula:

λ = h / √(2mK)

Where λ is the de Broglie wavelength, h is the Planck's constant, m is the mass of the proton, and K is the kinetic energy.

Given:

K = 50 keV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

h = 6.63 x 10^-34 J s

Substitute the values into the formula:

λ = 6.63 x 10^-34 / √(2 x 1.67 x 10^-27 x 80 x 10^-16)

λ ≈ 1.2826 x 10^-13 m

Therefore, the de Broglie wavelength for a proton with energy 50 keV is approximately 1.2826 x 10^-13 meters, which is equivalent to 128.26 angstroms (x 10^-10 m) to 2 significant figures.