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

Grade 12ICSEPhysics

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

🔑Concepts

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Einstein proposed that light consists of discrete packets of energy called 'quanta' or 'photons'. The energy of each photon is given by E=hνE = h\nu, where hh is Planck's constant and ν\nu is the frequency.

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The Work Function (Φ0\Phi_0) is the minimum energy required by an electron to escape from the metal surface. It is a characteristic property of the metal.

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

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Einstein’s Photoelectric Equation: When a photon of energy hνh\nu strikes a metal, a part of its energy is used as work function Φ0\Phi_0 to liberate the electron, and the remaining part appears as the maximum kinetic energy (KmaxK_{max}) of the emitted photoelectron.

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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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Intensity vs. Frequency: The number of photoelectrons emitted per second (photoelectric current) is directly proportional to the intensity of incident radiation, while the maximum kinetic energy of photoelectrons depends only on the frequency of incident radiation, not 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=hν−Φ0K_{max} = \frac{1}{2}mv_{max}^2 = h\nu - \Phi_0

Kmax=h(ν−ν0)K_{max} = h(\nu - \nu_0)

eV0=hν−Φ0eV_0 = h\nu - \Phi_0

V0=(he)ν−Φ0eV_0 = \left(\frac{h}{e}\right)\nu - \frac{\Phi_0}{e}

💡Examples

Problem 1:

Light of frequency 8×1014 Hz8 \times 10^{14} \text{ Hz} is incident on a metal surface. If the work function of the metal is 2.5 eV2.5 \text{ eV}, find the maximum kinetic energy of the emitted photoelectrons. (Take h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s} and 1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J})

Solution:

  1. Convert energy of incident photon to eV: E=hν=(6.63×10−34 J s)×(8×1014 Hz)=5.304×10−19 JE = h\nu = (6.63 \times 10^{-34} \text{ J s}) \times (8 \times 10^{14} \text{ Hz}) = 5.304 \times 10^{-19} \text{ J}.
  2. In eV: E=5.304×10−191.6×10−19≈3.315 eVE = \frac{5.304 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.315 \text{ eV}.
  3. Using Einstein's equation: Kmax=E−Φ0=3.315 eV−2.5 eV=0.815 eVK_{max} = E - \Phi_0 = 3.315 \text{ eV} - 2.5 \text{ eV} = 0.815 \text{ eV}.

Explanation:

The maximum kinetic energy is calculated by subtracting the work function (energy lost to escape the metal) from the total energy of the incident photon.

Problem 2:

The threshold wavelength for a certain metal is 6000 A˚6000 \text{ \AA}. Calculate the stopping potential when light of wavelength 4000 A˚4000 \text{ \AA} is incident on it.

Solution:

  1. Work Function Φ0=hcλ0\Phi_0 = \frac{hc}{\lambda_0}.
  2. Incident Energy E=hcλE = \frac{hc}{\lambda}.
  3. eV0=hcλ−hcλ0=hc(1λ−1λ0)eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right).
  4. Using hc≈12400 eV A˚hc \approx 12400 \text{ eV \AA}: V0=12400(14000−16000) VoltsV_0 = 12400 \left( \frac{1}{4000} - \frac{1}{6000} \right) \text{ Volts}.
  5. V0=12400(3−212000)=1240012000≈1.033 VV_0 = 12400 \left( \frac{3-2}{12000} \right) = \frac{12400}{12000} \approx 1.033 \text{ V}.

Explanation:

Stopping potential V0V_0 is numerically equal to the maximum kinetic energy expressed in electron-volts. Here we use the relationship between wavelength and energy.

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