Question

An electron is accelerated by a potential difference of 50v, what is the de broglie wavelength?

Answer 1

`1.73553* 10^(-10)\ m`

Step-by-step explanation:

h = Planck's constant =
`6.626* 10^(-34)\ m^2kg/s`

K = Potential difference = 50 V

m = Mass of electron =
`9.11* 10^(-31)\ kg`

The de broglie wavelength is given by

`\lambda=(h)/(√(2mK))\\\Rightarrow \lambda=\frac{6.626* 10^(-34)}{\sqrt{2* 9.11* 10^(-31)* 50* 1.6* 10^(-19)}}\\\Rightarrow \lambda=1.73553* 10^(-10)\ m`

The wavelength is
`1.73553* 10^(-10)\ m`