Particle nature of light (The Photon)
Particle nature of light :- Photoelectric effect i.e. the ejection of electrons from a metal surface when light of suitable frequency falls on it, established the fact that light when interact with matter behaves as if it was made of quanta (packets of energy) each having energy hν, where h is Planck’s Constant and ν is the frequency of the light. Einstein already associated the light quantum with momentum(hν/c), where c is the velocity of light.
Now the quanta has a definite amount of energy(hν) as well as momentum(hν/c). These results strongly indicate that light Quantum behaves like a particle which was later named as photon.
Characteristic properties of photons
(Particle nature of light)
(1). Energy of each Photon is given by and momentum where h is Planck’s Constant and ν and λ are the frequency and wavelength of the radiation and c is the velocity of light.
(2). The energy and momentum of a photon depends on frequency or wavelength and not on the intensity of radiation, so all the photons of a particular frequency ν and wavelength λ have the same energy E and same momentum p.
(3). If λ is in Å, then .
(4). All photons travel with the same speed c (speed of light) in vacuum but as the medium changes, the speed (v) changes (due to change in wavelength λ).
(5). The frequency of the photon(ν) does not change when it travels through different media.
(6). The increase in the intensity of light only increases the number of photons crossing an area per unit time. It does not affect the energy of the radiation.
(7). As a photon travel with the speed of light(c), so using , we get rest mass of photon zero.
(8). Dynamic mass of photon is determined by Einstein’s Mass-Energy Equivalence, i.e.,
(9). A photon remains unaffected by electric and magnetic fields. It is electrically neutral.
(10). In a photon-particle collision, the total energy and total momentum are conserved. However, the number of photons may not be conserved in a collision.
Example 1.
The energy of a photon is 1 MeV. Calculate the momentum and wavelength of photon. Given that Planck’s Constant h = 6.62 × 10-34 Js, speed of light c = 3 × 108 m/s.
Solution.
Energy of photon,
So, wavelength of photon
⇒ λ = 1.24 × 10-12 m
⇒ λ = 0.0124 Å
Momentum of photon
⇒ p =5.33 × 10-22 Kg m/s
NCERT Example 11.1
Monochromatic light of frequency 6.0 ×1014 Hz is produced by a laser. The power emitted is 2.0 ×10–3 W. (a) What is the energy of a photon in the light beam? (b) How many photons per second, on an average, are emitted by the source?
Solution.
(a) Energy of each photon,
E = hν= (6.63 ×10–34) (6.0 ×1014 )
⇒ E = 3.98 × 10-19 J
(b) Let N be the number of photons emitted by the source per second, so the power P transmitted by the source P = N E. Then
⇒ N = 5 × 1015 photons/sec
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