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

Answer:

$\phi=1.55 [eV]$

Explanation:

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

$\phi=\frac{hc}{\lambda}-K_{max}$

• 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:

$\phi=\frac{hc}{\lambda}-e\Delta V$

$\phi=\frac{6.63*10^{-34}*3*10^{8}}{546.1*10^{-9}}-0.728eV$    

$\phi=3.64*10^{-19}[J]-0.728 [eV]$

$\phi=(3.64*10^{-19}[J]*\frac{1eV}{1.6*10^{-19}[J]})-0.728 [eV]$

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

$\phi=1.55 [eV]$

I hope it helps you!

Answer 2

Answer:

The work function is $\phi = 1.544eV$

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

          $KE_{max} = \frac{hc}{\lambda} - \phi$

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

            $\phi = \frac{hc}{\lambda} - KE_{max}$

Substituting values

           $\phi = \frac{4.1357 *10^{-15} * (3.0 *10^8)}{546.1 *10^{-9}} - 0.728$

            $\phi = 1.544eV$