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Dual Nature of Radiation and Matter - Particle Nature of Light: The Photon

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Light behaves as if it is composed of discrete packets of energy called quanta or photons.

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Each photon of frequency ν\nu and wavelength λ\lambda has energy E=hν=hcλE = h\nu = \frac{hc}{\lambda}, where hh is Planck's constant (6.63×10−34 J s6.63 \times 10^{-34} \text{ J s}).

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Photons travel with the speed of light c=3×108 m/sc = 3 \times 10^8 \text{ m/s} in a vacuum, regardless of the frame of reference.

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The rest mass of a photon is zero (m0=0m_0 = 0). However, it possesses an equivalent dynamic mass m=Ec2m = \frac{E}{c^2}.

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Photons carry linear momentum given by p=Ec=hλp = \frac{E}{c} = \frac{h}{\lambda}.

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Photons are electrically neutral and are not deflected by electric or magnetic fields.

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In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, the number of photons may not be conserved; a photon may be absorbed or a new one may be created.

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Intensity of light depends on the number of photons crossing a unit area per unit time (nn). Higher intensity means more photons per second, but the energy of each individual photon remains the same if the frequency is constant.

📐Formulae

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

p=Ec=hλp = \frac{E}{c} = \frac{h}{\lambda}

m=Ec2=hcλm = \frac{E}{c^2} = \frac{h}{c\lambda}

n=PE=Pλhcn = \frac{P}{E} = \frac{P \lambda}{hc}

E(in eV)≈1242λ(in nm)E(\text{in eV}) \approx \frac{1242}{\lambda (\text{in nm})}

💡Examples

Problem 1:

Calculate the energy of a photon of blue light with a wavelength of 450 nm450 \text{ nm}. Given h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s} and c=3×108 m/sc = 3 \times 10^8 \text{ m/s}. Express the result in electron-volts (1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}).

Solution:

E=hcλ=6.63×10−34×3×108450×10−9E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{450 \times 10^{-9}} E=4.42×10−19 JE = 4.42 \times 10^{-19} \text{ J} EeV=4.42×10−191.6×10−19≈2.76 eVE_{\text{eV}} = \frac{4.42 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.76 \text{ eV}

Explanation:

The energy is calculated using the Planck-Einstein relation. We convert the wavelength from nanometers to meters and then convert the resulting energy from Joules to eV by dividing by the elementary charge.

Problem 2:

A monochromatic laser source of power 2 mW2 \text{ mW} emits light of wavelength 600 nm600 \text{ nm}. Find the number of photons emitted per second.

Solution:

P=2×10−3 WP = 2 \times 10^{-3} \text{ W} E=hcλ=6.6×10−34×3×108600×10−9=3.3×10−19 JE = \frac{hc}{\lambda} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} = 3.3 \times 10^{-19} \text{ J} n=PE=2×10−33.3×10−19≈6.06×1015 photons/sn = \frac{P}{E} = \frac{2 \times 10^{-3}}{3.3 \times 10^{-19}} \approx 6.06 \times 10^{15} \text{ photons/s}

Explanation:

The power PP represents the total energy emitted per second. By dividing the total power by the energy of a single photon (EE), we obtain the photon flux or the number of photons emitted per unit time.