Photon energy from wavelength: 500 nm is 2.49 eV

Working

Step by step

ν = c / λ = 3 × 10⁸ / (500 × 10⁻⁹) = 6.0 × 10¹⁴ Hz
E = hν = 6.63 × 10⁻³⁴ × 6.0 × 10¹⁴ = 3.98 × 10⁻¹⁹ J = 3.98 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 2.49 eV
ν₀ = φ / h = 2.14 × 1.6 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.16 × 10¹⁴ Hz, so λ₀ = c / ν₀ = 581 nm
K max = hν - φ = 2.486 - 2.14 = 0.35 eV
eV₀ = K max, so V₀ = 0.35 V
v max = √(2K / m) = √(2 × 5.54 × 10⁻²⁰ / 9.11 × 10⁻³¹) = 3.49 × 10⁵ m/s
Photons per second = P / E = 2.0 × 10⁻³ W / 3.98 × 10⁻¹⁹ J = 5.0 × 10¹⁵

Across the spectrum

Wavelength	Frequency	Energy (J)	Energy (eV)
400 nm	7.5 × 10¹⁴ Hz	4.97 × 10⁻¹⁹ J	3.11 eV
450 nm	6.67 × 10¹⁴ Hz	4.42 × 10⁻¹⁹ J	2.76 eV
500 nm	6.0 × 10¹⁴ Hz	3.98 × 10⁻¹⁹ J	2.49 eV
550 nm	5.45 × 10¹⁴ Hz	3.62 × 10⁻¹⁹ J	2.26 eV
600 nm	5.0 × 10¹⁴ Hz	3.31 × 10⁻¹⁹ J	2.07 eV
650 nm	4.62 × 10¹⁴ Hz	3.06 × 10⁻¹⁹ J	1.91 eV
700 nm	4.29 × 10¹⁴ Hz	2.84 × 10⁻¹⁹ J	1.78 eV

Derivation

Light arrives in packets, photons, each of energy E = hν.
One photon is absorbed by one electron, all at once, so there is no delay however dim the light.
The electron must spend at least the work function φ to leave the metal.
The fastest electrons keep the rest: K max = hν - φ. This is Einstein's photoelectric equation.
A retarding voltage V₀ stops even these, so eV₀ = K max. When hν = φ, K max = 0: that frequency is the threshold ν₀ = φ / h.

Questions

How do you find the energy of a photon from its wavelength?

E = hc / λ, with h = 6.63 × 10⁻³⁴ J s and c = 3 × 10⁸ m/s and λ in metres. Divide by 1.6 × 10⁻¹⁹ to get electron volts.

Which has more energy, a red photon or a violet one?

Violet. E = hc / λ, so the shorter wavelength carries more energy: about 3.1 eV at 400 nm against 1.8 eV at 700 nm.

Can 500 nm light free electrons from caesium?

Yes. 2.49 eV is more than caesium's work function of 2.14 eV, so electrons leave with up to 0.35 eV.