Final answer:
On reducing the wavelength of a photon, both its energy (E) and momentum (P) will increase because energy is directly proportional to frequency, and as wavelength decreases, frequency increases.
Step-by-step explanation:
If E and P are the energy and momentum of a photon respectively, then on reducing the wavelength of the photon, both P (momentum) and E (energy) will increase. This is because the energy of a photon is given by the equation E = hf, where h is Planck's constant and f is the frequency of the photon. As the wavelength (λ) decreases, the frequency (f) increases since they are inversely related (c = λf where c is the speed of light). Therefore, the energy E of the photon increases with increasing frequency.
Similarly, the momentum of a photon is given by p = E/c, and since E increases as wavelength decreases, the momentum P will also increase. Hence, the photon with the shorter wavelength will have more energy and momentum than the one with the longer wavelength.