QUESTION:3 (Start on a new page.) A photon of wavelength 2times 10^-7 m is located within the x-ray range of the electromagnetic spectrum. 3.1 Write down the speed of this photon in a vacuum. (1) 3.2 Calculate: 3.2.1 The frequency of the photon. (3) 5.3.2 The energy of the photon. (3) 3.3 The wavelength of the photon is doubled Will the penetrating ability of this photon INCREA SE, DECREASE or REMAIN THE SAME? Explain the answer. (3)

3.1 The speed of a photon in a vacuum is always constant and equal to the speed of light, which is approximately $$3 \times 10^8$$ m/s.

3.2.1 The frequency of a photon can be calculated using the formula: frequency = speed of light / wavelength. Given that the speed of light is $$3 \times 10^8$$ m/s and the wavelength is $$2 \times 10^{-7}$$ m, the frequency can be calculated as follows:

frequency = $$3 \times 10^8$$ m/s / $$2 \times 10^{-7}$$ m = $$1.5 \times 10^{15}$$ Hz

3.2.2 The energy of a photon can be calculated using the formula: energy = frequency * Planck's constant. Given that the frequency is $$1.5 \times 10^{15}$$ Hz and Planck's constant is $$6.63 \times 10^{-34}$$ J*s, the energy can be calculated as follows:

energy = $$1.5 \times 10^{15}$$ Hz * $$6.63 \times 10^{-34}$$ J*s = $$9.945 \times 10^{-18}$$ J

3.3 If the wavelength of the photon is doubled, the penetrating ability of the photon will DECREASE. This is because the energy of a photon is inversely proportional to its wavelength. As the wavelength increases, the energy decreases, resulting in a decrease in the penetrating ability of the photon.