A photon is an elementary particle that has zero invariant mass and travels in a vacuum with a constant velocity. It is capable of generating electromagnetic phenomena since it carries all forms of electromagnetic radiation:
- X-rays
- Ultraviolet light
- Visible light
- Gamma rays
- Infrared light
- Radio waves
- Microwaves and radio waves
The photon presents wave and corpuscular properties, this means that it behaves as a wave in certain phenomena, such as refraction in a lens; and as a particle when it interacts with matter to transfer energy. The latter is expressed as follows:
[tex]E=\frac{hc}{lambda} = hv[/tex]
- E is the energy, in physics it is defined as the capacity to do work.
- h is Planck's constant, a physical constant that plays a role in the theory of quantum mechanics and it is a constant between the energy of a photon and the frequency of its electromagnetic wave ([tex]E = hv[/tex]) Its value is 6.63 x 10^-34 J/s.
- c is the speed of light, which is 3x10^8 m/s.
- lambda (λ) is the wavelength, the distance traveled by a periodic disturbance propagating in one cycle.
- v is the frequency of the wave, it is the inverse of the wavelength, the number of repetitions per unit time of any periodic event.
This means that each photon has energy that is proportional to the frequency of light. In this case, the energy is quantized, which means that the energy of the photons is restricted to certain values. The energy at subatomic levels occurs in packets that refer to the photon, which are the packages of energy and each one correspond to different types of radiation.
This differs from classical waves, which can gain or lose arbitrary amounts of energy.
If we apply the equation and replace the information, we have:
[tex]E=\frac{6.63 x 10^{-34} J/s x 3 x 10^{8} m/s }{498 nm} = 3.99 x 10^{-28} J[/tex]
So 3.99 x 10^-28 J is the amount of energy of the photon of 498 nm emmited from a silicon atom.
Now we will show that the atomic levels are quantized. An atom has ground state (n=1) energy of 13.6 eV. Higher states means the atom is at an excited state. When this happens, an electron from an atom loses energy and makes a transition to a lower state. To avoid this and conserve energy, the atom emits a photon with an amount of energy that equals the difference of energy between a ground state and another state. So we can calculate the energy of the electron in different states.
[tex]E_{n} = \frac{-13.6 ev}{n^{2} }[/tex]
[tex]E_{1} = \frac{-13.6 ev}{1^{2} }=-13.6 ev[/tex]
[tex]E_{2} = \frac{-13.6 ev}{2^{2} }=-3.4 ev[/tex]
[tex]E_{3} = \frac{-13.6 ev}{3^{2} }=-1.511 ev[/tex]
N represents the principal quantum number, which is the overall energy of each orbital. This energy increases as its distance from the nucleus increases. Here we can se that the value of energy of n=1 is higher than energy of n=2, and this one is higher from the energy of n=3. This means each level has a specific energy and it is quantized according to the value of n.
So, a photon is an elementary particle which carries radiation and presents a wave-corpuscle duality, where particles may exhibit wave-like behavior in some experiments while appearing as particles in others. A photon is emmited when an atom loses energy, to conserve this energy and avoid a transition to a lower state. The electron energy is quantized because it is restricted to characteristic values, only taking non-continuous values. In this question, the amount of energy of the photon emmited is 3.99 x 10^-28 J. And the atomic levels are quantized according to the principal quantum number.
Learn more about photons here: https://brainly.com/question/18101137
