Question

in photoelectric effect how are the intensity ,frequency ,stopping potential are related and why

Answer 1

Einstein equation: Energy= W+Ek, where Ek=(mv^2)/2=eU
U-stopping potential,W-work function which is W=hf0, f0-frequence.

Answer 2

The Einstein equation uses Plank's concept of that light delivers energy to an absorber in quanta of energy
\[h\nu\]
to say that given the amount of energy it takes to lift an electron out of a surface (the work function), its kinetic energy can be shown to be
\[ K = h\nu - \phi\]
K is the kinetic energy of an ejected electron from a surface (a metal in our case)
h is Plank's constant
nu is the frequency of the light striking the electron
phi is the work function, an experimentally found value. It can be written like @Klodi had as
\[\phi = h\nu_o\]
the nu_o is the minimum frequency of light striking the metal surface for an electron to gain enough kinetic energy to be emitted. Writing the equation this way
\[ k = h(\nu - \nu_o)\]

shows the basis of the experiment - that an electron has to be imparted with enough energy through excitation by a photon of energy of AT LEAST hf_o to escape the surface of a metal, and does so in a discrete fashion proportional to Plank's constant.

There is no intensity in this equation. Intensity of the light has an effect on the number of electrons being emitted, but not on the energy at which they are emitted due to the quantized nature of electron emission. So even if the intensity of the radiation (ligh) source is immense (up to a point), if the frequency of its photons is not great enough, no photoelectrons will be emitted.

As for the stopping potential, we can that the above equation can be written as
\[eV_{sp} = h\nu - \phi \]

by knowing that: since the emitted electrons have a maximum kinetic energy for any given constant frequency and intensity of radiation (light), that energy can be calculated as

\[K_{max} = \frac{1}{2}mv_{max}^2\]

so If we set that equal to a potential energy, U = qV, with our q=e- then
\[K_{max} = eV_{sp}\]


Rearranging, we can see that by holding constant a stopping potential

\[V_{sp} = h\frac{ \nu}{e} - \frac{\phi}{e} \]
we can come up with line of the basic form y = mx+b and experimentally find a value for Plank's constant as a function of the radiation (light) frequency wherein h is the slope of the line.

Answer 3

~~
\[\frac{h}{e} \] would be the slope of the line. derp.

Answer 4

that becomes less useful much more quickly than I had thought...

The intensity dictates the number of electrons lifted from a surface for a given frequency. Frequency dictates if an electron will absorb enough energy to be lifted from a surface. The stopping potential is a parameter that is defined by the maximum kinetic energy of an escaping electron given the Einstein equation.

Answer 5

thanks