Question

In a photoelectric experiment, a metal is irradiated with light of energy 3.56 eV. If a stopping potential of 1.10 V is required, what is the work function of the metal?

Answer 1

The work function is
`\phi = 2.46 \ eV`

Step-by-step explanation:

From the question we are told that

The light energy is
`E = 3.56 eV`

The stopping voltage is
`V = 1.10 \ V`

Generally work function is mathematically represented as

`\phi = E - KE`

Where KE is the kinetic energy of the ejected electron and it is mathematically represented as

`KE = V * e`

Where e is the charge on the electron

So

`KE = 1.10eV`

Thus

`\phi = 3.56eV - 1.10 eV`

=>
`\phi = 2.46 \ eV`