Question

Question #8711a

Answer 1

This question is an example of the photoelectric effect

Photoelectric Effect

When a photon of sufficient energy hits the surface of an object electron are released depending on the energy of the photon.Less electrons are emitted when the object is hit by a photon of less energy and vice versa.

The stopping potential is the potential energy measured in volts (joules/coulomb) that must be applied to stop the electrons from being ejected from the surface when the light is shone on it.

This will make the KE of the electron 0.

`V_s = (h c) / (λ q e) - ϕ`

Plug in the variables

`2.15V = ((6.63 xx10^ − 34Js)∗ (3 xx10^8"m/s") )/((200xx10^-9 m)(1.6*10^-19C)) - x`

`2.15V = 6.215625 - x`

`x = 6.215625 - 2.15V = 4.065625eV`

4.065625eV is the work function of cadium