Question

derivation of stoping potential​

Answer 1

Answer:

Einstein and Millikan described the photoelectric effect using a formula (in contemporary notation) that relates the maximum kinetic energy (Kmax) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f0) of the photoemissive surface.

Kmax = h(f − f0)

or if you prefer, to the energy of the absorbed photons (E) and the work function (ϕ) of the surface

Kmax = E − ϕ

where the first term is the energy of the absorbed photons (E) with frequency (f) or wavelength (λ)

E = hf =  hc

λ

and the second term is the work function (ϕ) of the surface with threshold frequency (f0) or threshold wavelength (λ0)

ϕ = hf0 =  hc

λ0

The maximum kinetic energy (Kmax) of the photoelectrons (with charge e) can be determined from the stopping potential (V0).

V0 =  W  =  Kmax

q e

Thus…

Kmax = eV0

When charge (e) is given in coulombs, the energy will be calculated in joules. When charge (e) is given in elementary charges, the energy will be calculated in electron volts. This results in a lot of constants. Use the one that's most appropriate for your problem.

Planck constant with variations

 SI units acceptable

non SI units

h 6.63 × 10−34 Js 4.14 × 10−15 eVs

hc 1.99 × 10−25 Jm 1240 eVnm

Lastly, the rate (n/t) at which photoelectrons (with charge e) are emitted from a photoemissive surface can be determined from the photoelectric current (I).

I =  q  =  ne

t t

Thus…

n  =  I

t e

Explanation: