Question

Find the retarding potential required to stop electron of the de broglie wavelength 0.5 nm.

Answer 1

use the formula de broglie wavelength=√(150/v) in Angstrom.
V is the potential.

Answer 2

Answer:

V = 6.02 volts

Explanation:

Given that,

Wavelength of the electron, $\lambda=0.5\ nm=0.5\times 10^{-9}\ m$

Let V is the retarding potential required to stop electron. The relation between the potential and the wavelength is given by :

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

V is the potential

Putting all the known values in above formula,

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}V}}$

$\lambda=\dfrac{1.227}{\sqrt{V}}\times 10^{-9}$

$0.5\times 10^{-9}=\dfrac{1.227}{\sqrt{V}}\times 10^{-9}$

V = 6.02 volts

So, the retarding potential required to stop electron is 6.02 volts. Hence, this is the required solution.