To determine which color of light has the maximum energy associated with its photons, we need to consider the relationship between the energy of a photon and its wavelength. The energy \((E)\) of a photon is given by the equation:
\(E = \frac{hc}{\lambda}\)
where:
From the equation, we can see that the energy of a photon is inversely proportional to its wavelength. This means that the shorter the wavelength, the higher the energy of the photon.
Let's consider the visible spectrum and the typical wavelength range for each color:
Among the given options, blue light has the shortest wavelength. Therefore, according to the inverse relationship between wavelength and energy, photons of blue light have the maximum energy compared to the other options listed.
Hence, the correct answer is Blue light.
Rather than comparing wavelengths, this can be reasoned directly through frequency, since a photon's energy is \( E = h\nu \), and frequency and wavelength are inversely related through \( \nu = \dfrac{c}{\lambda} \), the colour with the highest frequency of visible light will have the highest photon energy.
Across the visible spectrum, frequency rises steadily as we move from red toward violet: red light has the lowest frequency of the visible colours, followed by orange, yellow, green, blue, and finally violet at the highest frequency.
Among the four options given, red has by far the lowest frequency, yellow is only slightly higher, green is higher still, and blue sits at the highest frequency of the group, closest to the violet end of the spectrum.
Since \( E = h\nu \) increases in direct proportion with frequency, and blue light has the greatest frequency among the four choices, its photons must carry the greatest energy.
Therefore, the correct answer is Blue light.