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Dual Nature of Radiation and Matter - Einstein’s Photoelectric Equation

Grade 12CBSEPhysics

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

🔑Concepts

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According to Einstein’s photon theory, light of frequency ν\nu consists of discrete packets of energy called photons, where each photon has energy E=hνE = h\nu.

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When a photon of energy hνh\nu strikes a metal surface, it is completely absorbed by a single electron. This energy is used to overcome the work function ϕ0\phi_0 of the metal and the remainder is given as maximum kinetic energy KmaxK_{max} to the electron.

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The Work Function ϕ0\phi_0 is defined as the minimum energy required by an electron to escape from the metal surface, represented as ϕ0=hν0\phi_0 = h\nu_0, where ν0\nu_0 is the threshold frequency.

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The maximum kinetic energy KmaxK_{max} of the emitted photoelectrons is independent of the intensity of incident light but depends linearly on the frequency ν\nu of the incident radiation.

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The stopping potential V0V_0 is the negative potential applied to the collector plate at which the photoelectric current becomes zero. It is related to maximum kinetic energy by Kmax=eV0K_{max} = eV_0.

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The photoelectric effect is an instantaneous process, occurring within 10−9 s10^{-9} \text{ s} or less, provided ν>ν0\nu > \nu_0.

📐Formulae

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

Kmax=hν−ϕ0K_{max} = h\nu - \phi_0

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

eV0=hν−hν0eV_0 = h\nu - h\nu_0

u_0)$$

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

💡Examples

Problem 1:

Light of frequency 8.41×1014 Hz8.41 \times 10^{14} \text{ Hz} is incident on a metal surface. Electrons with a maximum speed of 0.6×106 m/s0.6 \times 10^6 \text{ m/s} are ejected. Calculate the threshold frequency for the metal. (Take h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s} and m=9.1×10−31 kgm = 9.1 \times 10^{-31} \text{ kg})

Solution:

  1. Calculate KmaxK_{max}: Kmax=12mv2=12×(9.1×10−31)×(0.6×106)2=1.638×10−19 JK_{max} = \frac{1}{2}mv^2 = \frac{1}{2} \times (9.1 \times 10^{-31}) \times (0.6 \times 10^6)^2 = 1.638 \times 10^{-19} \text{ J}.
  2. Use Einstein's equation: hν=ϕ0+Kmax  ⟹  hν=hν0+Kmaxh\nu = \phi_0 + K_{max} \implies h\nu = h\nu_0 + K_{max}.
  3. Rearrange for ν0\nu_0: ν0=ν−Kmaxh=(8.41×1014)−1.638×10−196.63×10−34=8.41×1014−2.47×1014=5.94×1014 Hz\nu_0 = \nu - \frac{K_{max}}{h} = (8.41 \times 10^{14}) - \frac{1.638 \times 10^{-19}}{6.63 \times 10^{-34}} = 8.41 \times 10^{14} - 2.47 \times 10^{14} = 5.94 \times 10^{14} \text{ Hz}.

Explanation:

The threshold frequency ν0\nu_0 is found by subtracting the frequency corresponding to the kinetic energy from the incident frequency.

Problem 2:

The work function of cesium is 2.14 eV2.14 \text{ eV}. Find the stopping potential for the photoelectrons emitted when light of frequency 6×1014 Hz6 \times 10^{14} \text{ Hz} is incident on the metal surface.

Solution:

  1. Convert frequency to energy in eV: E=hν=6.63×10−34×6×10141.6×10−19≈2.48 eVE = h\nu = \frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}} \approx 2.48 \text{ eV}.
  2. Apply Einstein's equation: Kmax=E−ϕ0=2.48 eV−2.14 eV=0.34 eVK_{max} = E - \phi_0 = 2.48 \text{ eV} - 2.14 \text{ eV} = 0.34 \text{ eV}.
  3. Since Kmax=eV0K_{max} = eV_0, the stopping potential V0=0.34 VV_0 = 0.34 \text{ V}.

Explanation:

The stopping potential in Volts is numerically equal to the maximum kinetic energy expressed in electron-volts (eV).

Einstein’s Photoelectric Equation - Revision Notes & Key Formulas | CBSE Class 12 Physics