Concept:
According to Planck's quantum theory, the energy of a photon is given by
\[
E=h\nu,
\]
where
• \(h\) is Planck's constant,
• \(\nu\) is the frequency of radiation.
Since
\[
\nu=\frac{c}{\lambda},
\]
the energy can also be written as
\[
E=\frac{hc}{\lambda}.
\]
Thus, photon energy is inversely proportional to wavelength.
\[
E\propto \frac{1}{\lambda}.
\]
Hence, the colour having the smallest wavelength will possess the highest photon energy.
Step 1: Recall the order of visible colours according to wavelength.
The visible spectrum follows the order
\[
\text{Red} \rightarrow \text{Orange} \rightarrow \text{Yellow}
\rightarrow \text{Green} \rightarrow \text{Blue}
\rightarrow \text{Violet}.
\]
As we move from red to violet:
• Wavelength decreases.
• Frequency increases.
• Photon energy increases.
Step 2: Compare the given colours.
Among the options:
\[
\lambda_{\text{blue}}
<
\lambda_{\text{green}}
<
\lambda_{\text{yellow}}
<
\lambda_{\text{red}}.
\]
Therefore,
\[
E_{\text{blue}}
>
E_{\text{green}}
>
E_{\text{yellow}}
>
E_{\text{red}}.
\]
Step 3: Identify the colour having maximum photon energy.
Since blue light has the highest frequency among the given options,
\[
E=h\nu
\]
will be maximum for blue light.
Thus,
\[
\boxed{\text{Blue light}}
\]
has the maximum photon energy.
Hence, the correct answer is
\[
\boxed{\text{(D)}}
\]