The equation for the energy of a photon is

$$ E = hf $$

Where \(h\) is Planck's constant

$$ h = 6.6 \times 10^{-34} \text{ Js} $$

So for a photon of wavelength

$$ \lambda = 650 \text{ nm} $$

$$ f = \frac{c}{\lambda} $$

$$ f = \frac{3 \times 10^8}{6.5 \times 10^{-7}} $$

$$ f = 4.62 \times 10^{14} \text{ Hz}$$

So

$$ E = hf $$

$$ E = 6.6 \times 10^{-34} \times 4.62 \times 10^{14} $$

$$ E = 3.05 \times 10^{-19} J $$

Given the range in visible wavelengths is 700 nm to 400 nm which is a good approximation (NASA put lower limit at 380 nm) then we can calculate the range of photon energies for visible light.

$$ \lambda = 700 \text{ nm} $$

$$ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{7.0 \times 10^{-7}} $$

$$ E = 2.82 \times 10^{-19} J $$

and

$$ \lambda = 400 \text{ nm} $$

$$ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4.0 \times 10^{-7}} $$

$$ E = 4.95 \times 10^{-19} J $$

In terms of electron volts which is a more convenient unit of energy

Energies range from

$$ \text{red photon } 1.76 \text{ eV to blue photon} 3.09 \text{ eV}$$

In the photoelectric effect Einstein argued that the electron absorbs the photon's energy and some of it releases the electron - the work function - the remainder is the kinetic energy of the electron.

$$ hf = \phi + \text{KE}_{max} $$

So for an electron to be emitted at all

$$ hf = \phi $$

and

$$ f_{threshold} = \frac{\phi}{h} $$

If zinc has a work function of 4.33 eV find the maximum kinetic energy of the electron emitted it light of wavelenght 250 nm is incident on zinc.

$$ hf = \phi + \text{KE}_{max} $$

$$ \text{KE}_{max} = hf - \phi $$

$$ \text{KE}_{max} = h\frac{c}{\lambda} - \phi $$

$$ \text{KE}_{max} = 6.6 \times 10^{-34} \frac{3 \times 10^8}{2.5 \times 10^{-7}} - 4.33 \times 1.6 \times 10^{-19} $$

$$ \text{KE}_{max} = 7.9 \times 10^{-19} - 6.9 \times 10^{-19} $$

$$ \text{KE}_{max} = 1.0 \times 10^{-19} \text{ J} $$

If zinc has a work function of 4.33 eV find the threshold frequency for a photoelectron to just be emittd from the surface of zinc.

$$ hf = \phi + \text{KE}_{max} $$

but at threshold kinetic energy is zero. So

$$ hf = \phi $$

$$ hf = 4.33 \times 1.6 \times 10^{-19} $$

$$ hf = 6.9 \times 10^{-19} $$

$$ f = \frac{1.0 \times 10^{-19}}{6.6 \times 10^{-34}} $$

$$ f = 1.5 \times 10^{15} \text{ Hz} $$