Question

In a photoelectric experiment it is found that a stopping potential of 1.00 V is needed to stop all the electrons when incident light of wavelength 296 nm is used and 2.9 V is needed for light of wavelength 207 nm. From these data determine Planck's constant and the work function of the metal.

Answer 1

Knowing that the stopping potential is 1.00 V when the wavelength of the incident light is 296 nm and 2.9 V when the wavelength is 207 nm Planck's constant (h) and the work function (
`\phi`) of the metal are
`h=6,9904x10^(-34) J.s` and
`\phi=3,42 eV`.

Step-by-step explanation:

If an electron of a metal surface absorbs the energy of a photon and gets more energy than the binding energy of the metal (given by the work function) then is ejected.

The next equation describes the maximum kinetic energy of the ejected electrons:

`eV_(stop)=h\\u-\phi`

where

`e=1-1,6021766x10^(-19)C` is the charge of the electron

`\\u=(c)/(\lambda)` is the frequency of the incident light

`V_(stop)` is the stopping potential

to determine Planck's constant (
`h`) and the work function (
`\phi`) given the stopping potentials measured at two different wavelengths we have two equations:

`eV_(stop1) =h(c)/(\lambda_(1) ) -\phi\\\\eV_(stop2) =h(c)/(\lambda_(2) ) -\phi`

Planck's constant (
`h`)

We use that the speed of light is
`c=299792458(m)/(s)` and that

`V_(stop1)=1.00V \\\lambda_(1) =296nm=296x10^(-9) m\\V_(stop2)=2.9V \\\lambda_(1) =207nm=207x10^(-9) m\\`

and we replace these values in the equations to get

`1.00V=h(6,321x10^(33) (1)/(Cs) )-(\phi)/(1,6021766x10^(-19)C ) \\\\2.9V=h(9,039x10^(33) (1)/(Cs) )-(\phi)/(1,6021766x10^(-19)C )`

if we substract the second equation with the first one we get that:

`1.9V=h(2,718x10^(33)(1)/(Cs))\\\\h=6,9904x10^(-34) J.s`

We need to note that this number for Planck's constant is a little over the exact value
`h=6,62607x10^(-34) J.s`.

Work function (
`\phi`)

Then replacing the value we found for
`h` in one of the equations relating the stopping potential with the wavelength we get:

`\phi=h(c)/(\lambda_(1)) -eV_(stop1) \\\\\phi=(6,9904x10^(-34) J.s\quad 299792458(m)/(s))/(296x10^(-9)m) -1.00eV\\\\\phi=7,07996x10^(-19) J-1.00eV`

we need to make a little change of units to match the two terms in the right side of the equation, so we use
`1J=6,2415x10^(18) eV`

`\phi=4,42eV-1.00eV`

we finally get for the work function

`\phi=3,42eV`