A photon has twice the frequency of another photon with frequency 4.0 × 10^14 Hz. What is the energy of the higher-frequency photon?

What the problem is asking

You are told one photon’s frequency, then asked for the energy of a second photon whose frequency is twice as large. The link between frequency and photon energy is Planck’s relation.

Double the frequency first

Given $f_1 = 4.0\times10^{14}\,\text{Hz}$, the other photon has

$$f_2 = 2f_1 = 2(4.0\times10^{14}) = 8.0\times10^{14}\,\text{Hz}.$$

Convert frequency to energy with Planck’s constant

Use $E = hf$, where $h = 6.626\times10^{-34}\,\text{J·s}$:

$$E = (6.626\times10^{-34})(8.0\times10^{14}) = 5.3008\times10^{-19}\,\text{J} \approx 5.30\times10^{-19}\,\text{J}.$$

Optional: express the energy in electronvolts

Using $1\,\text{eV} = 1.602\times10^{-19}\,\text{J}$,

$$E = \frac{5.30\times10^{-19}}{1.602\times10^{-19}} \approx 3.31\,\text{eV}.$$

So the photon’s energy is $5.30\times10^{-19}\,\text{J}$ (about $3.31\,\text{eV}$).