The residual energy (h ν- φ0) emerges as electron kinetic energy. If the electron does not lose any of its energy in impact with the surface and exits with the greatest possible kinetic energy.
∴ 1 / 2 m v2max = hν – φ0 …..(1)
where,
- m = mass of the electron
- vmax = maximum velocity of electron
All the photoelectrons emitted from the metal surfaces do not have the same energy.
The photoelectric current becomes zero when the stopping potential is sufficient to repel even the most energetic photoelectrons, with the maximum kinetic energy, so that the stopping potential (in volts) is numerically equal to the maximum kinetic energy of photo-electron in eV.
1 / 2 m v2max = ev0 …..(2)
where,
- e = magnitude of the charge on an electron.
From (2) we have,
1 / 2 m v2max = hν – φ0
eV0 = hν – φ0
This equation is Einstein`s photoelectric equation.
With the aid of Einstein’s photoelectric equation, we can now describe all of the features of the photoelectric effect:
If the frequency of incident radiation is decreased, the kinetic energy of photoelectrons also decreases, and finally, it becomes zero for a particular frequency (say ν). ν is called the threshold frequency. Thus,
When ν=νo,
then
KEmax = 1/2 m V2max = 0
Therefore from equation (1), we get
0 = hν0 – φ0
hν0 = φ0
Therefore, Einstein`s equation can be written as,
1/2 m v2max = h(ν – ν0) …..(3)
From the above equation, we can say three points,
- ν > ν0, the photoelectrons are emitted with some velocity,
- ν < ν0, no photoelectrons are emitted, and
- ν = ν0, photoelectrons are just emitted with zero kinetic energy.
A more intense beam, according to quantum theory, includes a higher number of photons. As a result, the number of photon-electron collisions increases, and more photoelectrons are released. This explains why the photoelectric current increases with incoming photon intensity. Because the photoelectric work function (φ0) is constant for every given emitter, equation (3) demonstrates that the KEmax of photoelectrons grows with the frequency of incoming radiation but does not rely on intensity.
vmax = 6 × 105 m/s
ν0 = 4.22 × 1014 Hz
KEmax = 1/2 mv2max = h(ν – ν0)
ν = 1/2 × (mv2max / h) + ν0
= (1/2) × [(9.1 × 10-31 ×(6 × 105)2 / 6.63 × 10-34] +4.22 × 1014 Hz
= 2.47 × 1014 + 4.22 × 1014
= 6.69 × 1014 Hz
λ1 = 3310 = 3.31 × 10-7 m
KEmax (1) = 3 × 10-19 J
λ2 = 5000 = 5 × 10-7 m
KEmax = 9.72 × 10-20 J= 0.972 × 10-19 J
KEmax (1) = hc/λ1 – φ0
KEmax (2) = hc/λ2 – φ0
KEmax (1) – KEmax (2) = hc (1/λ1 – 1/λ2)
3 × 10-19 – 0.972 × 10-19 = h × 3 × 108 [(1/3.31 × 10-7) – (1/5 × 10-7)]
2.028 × 10-19 = h × 3 × 1015 (1.69 / 16.55)
h = 2.028 × 10-19 × 16.55 / 3 × 1015 × 1.69
= 6.62 × 10-34 Js
Now,
KEmax (1) = hc/λ1 – φ0
φ0 = hc/λ1 – KEmax (1)
= (6.62 × 10-34 × 3 × 108 / 3.31 × 10-7 ) – 3 × 10-19
= 3 × 10-19 J
Also,
φ0 = hc / λ0
λ0 = hc / φ0
= 6.62 × 10-34 × 3 × 108 /3 × 10-19
= 6.62 × 10-7
= 6620 A°