Question

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of the electron. (b) Would such an electron be useful in a Davisson-Germer type scattering experiment

Answer 1

λ = 6.19*10^-11 m

Step-by-step explanation:

Given that

Kinetic Energy, K = 390 keV

Mass of electron, m = 9.1*10^-31 kg

Planck's constant, h = 6.6*10^-34

To find the De Broglie's wavelength, we use this formula

λ = h / √(2mk), where λ is the wavelength were looking for.

Now we substitute the values into the equation

λ = 6.6*10^-34 / √(2 * 9.1*10^-31 * 390 * 1.6*10^-19)

λ = 6.6*10^-34 / √1.136*10^-46

λ = 6.6*10^-34 / 1.066*10^-23

λ = 6.19*10^-11 m