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Dual Nature of Radiation and Matter - Photoelectric Effect

Grade 12CBSEPhysics

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

🔑Concepts

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Photoelectric Effect: The phenomenon of emission of electrons from a metal surface when electromagnetic radiation of suitable frequency (above a threshold) falls on it.

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Photon: Electromagnetic radiation behaves like particles called photons. Each photon carries energy E=hνE = h\nu and momentum p=hλp = \frac{h}{\lambda}.

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Work Function (ϕ0\phi_0): The minimum energy required by an electron to just escape from the metal surface. It depends on the nature of the metal and its surface properties.

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Threshold Frequency (ν0\nu_0): The minimum frequency of incident radiation below which no photoelectric emission occurs, regardless of intensity.

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Stopping Potential (V0V_0): The minimum negative (retarding) potential applied to the anode for which the photoelectric current becomes zero. It is a measure of the maximum kinetic energy of the emitted electrons: Kmax=eV0K_{max} = eV_0.

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Einstein's Photoelectric Equation: It is based on the conservation of energy. Total energy of incident photon = Work function + Maximum kinetic energy of photoelectron (hν=ϕ0+Kmaxh\nu = \phi_0 + K_{max}).

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Intensity and Photoelectric Current: For a frequency above ν0\nu_0, the photoelectric current is directly proportional to the intensity of incident radiation.

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Independence of KmaxK_{max}: The maximum kinetic energy of photoelectrons depends on the frequency of incident radiation but is independent of its intensity.

📐Formulae

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

ϕ0=hν0=hcλ0\phi_0 = h\nu_0 = \frac{hc}{\lambda_0}

Kmax=12mvmax2=eV0K_{max} = \frac{1}{2}mv_{max}^2 = eV_0

hν=ϕ0+eV0h\nu = \phi_0 + eV_0

eV0=h(ν−ν0)eV_0 = h(\nu - \nu_0)

V0=(he)ν−ϕ0eV_0 = \left( \frac{h}{e} \right)\nu - \frac{\phi_0}{e}

p=hλ=Ecp = \frac{h}{\lambda} = \frac{E}{c}

💡Examples

Problem 1:

Monochromatic light of frequency 6.0×1014 Hz6.0 \times 10^{14}\text{ Hz} is produced by a laser. The power emitted is 2.0×10−3 W2.0 \times 10^{-3}\text{ W}. (a) What is the energy of a photon in the light beam? (b) How many photons per second, on average, are emitted by the source?

Solution:

(a) Energy of a photon E=hν=(6.63×10−34 J s)×(6.0×1014 Hz)=3.98×10−19 JE = h\nu = (6.63 \times 10^{-34}\text{ J s}) \times (6.0 \times 10^{14}\text{ Hz}) = 3.98 \times 10^{-19}\text{ J}. (b) Number of photons per second n=PE=2.0×10−3 W3.98×10−19 J≈5.0×1015 photons/sn = \frac{P}{E} = \frac{2.0 \times 10^{-3}\text{ W}}{3.98 \times 10^{-19}\text{ J}} \approx 5.0 \times 10^{15}\text{ photons/s}.

Explanation:

The energy of a single photon is determined by its frequency using Planck's constant. The total power is the energy delivered per second, so dividing power by the energy of one photon gives the photon flux.

Problem 2:

The work function of cesium is 2.14 eV2.14\text{ eV}. Find (a) the threshold frequency for cesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V0.60\text{ V}.

Solution:

(a) ν0=ϕ0h=2.14×1.6×10−19 J6.63×10−34 J s=5.16×1014 Hz\nu_0 = \frac{\phi_0}{h} = \frac{2.14 \times 1.6 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = 5.16 \times 10^{14}\text{ Hz}. (b) Using eV0=hcλ−ϕ0eV_0 = \frac{hc}{\lambda} - \phi_0, we get hcλ=eV0+ϕ0=0.60 eV+2.14 eV=2.74 eV\frac{hc}{\lambda} = eV_0 + \phi_0 = 0.60\text{ eV} + 2.14\text{ eV} = 2.74\text{ eV}. Thus, λ=hc2.74 eV=1242 eV nm2.74 eV≈453 nm\lambda = \frac{hc}{2.74\text{ eV}} = \frac{1242\text{ eV nm}}{2.74\text{ eV}} \approx 453\text{ nm}.

Explanation:

Threshold frequency is calculated directly from the work function. For the second part, the stopping potential gives the maximum kinetic energy in electron-volts, which is added to the work function to find the total incident photon energy.

Photoelectric Effect - Revision Notes & Key Formulas | CBSE Class 12 Physics