Question

A photoelectric experiment is performed where green light with a wavelength of 546.1 nm is shined on a metal plate, creating a photocurrent from it to a collector plate. When the potential difference between the metal plate and the collector is increased to a magnitude of 0.728 V, the photocurrent goes to zero—in other words, this is the stopping potential. What is the work function (in eV) for this metal?

Answer 1

The work function is
`\phi = 1.544eV`

Step-by-step explanation:

From the question we are told that

The wavelength of the green light is
`\lambda = 546.1nm`

The stopping potential is
`V= 0.728V`

At stopping potential the kinetic is also maximum because this where the electron(causing the current ) would flow at it highest speed before return to zero

So the maximum kinetic energy of this electron in term of electron volt is

`KE_(max) = 0.728 eV`

This maximum kinetic energy is mathematically represented as

`KE_(max) = (hc)/(\lambda) - \phi`

Where h is the planck's constant with a value
`h = 4.1357 *10^(-15) eV`

c is the speed of light
`c = 3.0 *10^8 m/s`

`\phi` is the work function

Making
`\phi` the subject of the formula

`\phi = (hc)/(\lambda) - KE_(max)`

Substituting values

`\phi = (4.1357 *10^(-15) * (3.0 *10^8))/(546.1 *10^(-9)) - 0.728`

`\phi = 1.544eV`

Answer 2

`\phi=1.55 [eV]`

Step-by-step explanation:

We can use the work function equation for a photoelectric experiment:

`\phi=(hc)/(\lambda)-K_(max)`

• h is the plank constant
• c is the speed of light
• λ is the wave length
• K is the kinetic energy (or K=eΔV)

So we will have:

`\phi=(hc)/(\lambda)-e\Delta V`

`\phi=(6.63*10^(-34)*3*10^(8))/(546.1*10^(-9))-0.728eV`

`\phi=3.64*10^(-19)[J]-0.728 [eV]`

`\phi=(3.64*10^(-19)[J]*(1eV)/(1.6*10^(-19)[J]))-0.728 [eV]`

`\phi=2.28 [eV] - 0.728 [eV]`

`\phi=1.55 [eV]`

I hope it helps you!