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Dual Nature of Radiation and Matter - Photoelectric Effect

Grade 12ICSEPhysics

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

🔑Concepts

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The phenomenon of emission of electrons from a metal surface when light of sufficiently high frequency falls on it is called the Photoelectric Effect. The emitted electrons are known as photoelectrons.

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The Work Function (Φ0\Phi_0 or W0W_0) is the minimum energy required by an electron to escape from a metal surface. It is characteristic of the metal and is measured in electron-volts (1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}).

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Threshold Frequency (ν0\nu_0) is the minimum frequency of incident radiation below which no photoelectrons are emitted, regardless of the intensity of light.

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

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Einstein's Photoelectric Equation: Kmax=hν−Φ0K_{max} = h\nu - \Phi_0. This suggests that the energy of an incident photon (hνh\nu) is used in two ways: to overcome the work function and to provide kinetic energy to the electron.

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Laws of Photoelectric Effect: (i) Photoelectric current is directly proportional to the intensity of light. (ii) Maximum kinetic energy depends on the frequency of incident light, not its intensity. (iii) The process is instantaneous (10−9 s10^{-9} \text{ s} or less).

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Dual Nature: Light behaves as a wave (interference, diffraction) and as a particle (photoelectric effect, Compton effect). Matter also exhibits this duality through de-Broglie waves.

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de-Broglie Wavelength: Any moving particle of mass mm and velocity vv has a wavelength associated with it, given by λ=hp\lambda = \frac{h}{p}.

📐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=hν0+Kmaxh\nu = h\nu_0 + K_{max}

eV0=hν−Φ0eV_0 = h\nu - \Phi_0

λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}

λ=h2mK=h2mqV\lambda = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}

λelectron≈12.27V A˚\lambda_{electron} \approx \frac{12.27}{\sqrt{V}} \text{ \AA}

💡Examples

Problem 1:

The work function of cesium is 2.14 eV2.14 \text{ eV}. Find the threshold frequency for cesium and the maximum kinetic energy of the photoelectrons emitted when light of frequency 6.0×1014 Hz6.0 \times 10^{14} \text{ Hz} is incident on the metal surface.

Solution:

Given: Φ0=2.14 eV=2.14×1.6×10−19 J\Phi_0 = 2.14 \text{ eV} = 2.14 \times 1.6 \times 10^{-19} \text{ J}. (i) Threshold Frequency: ν0=Φ0h=2.14×1.6×10−196.63×10−34≈5.16×1014 Hz\nu_0 = \frac{\Phi_0}{h} = \frac{2.14 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.16 \times 10^{14} \text{ Hz}. (ii) Maximum Kinetic Energy: Kmax=hν−Φ0=(6.63×10−34×6.0×1014)−(2.14×1.6×10−19)≈0.34 eVK_{max} = h\nu - \Phi_0 = (6.63 \times 10^{-34} \times 6.0 \times 10^{14}) - (2.14 \times 1.6 \times 10^{-19}) \approx 0.34 \text{ eV}.

Explanation:

We use the definition of work function to find threshold frequency and Einstein's equation to find the kinetic energy of the emitted photoelectrons.

Problem 2:

Calculate the de-Broglie wavelength associated with an electron accelerated through a potential difference of 100 V100 \text{ V}.

Solution:

Using the shortcut formula for electrons: λ=12.27V A˚\lambda = \frac{12.27}{\sqrt{V}} \text{ \AA}. λ=12.27100=12.2710=1.227 A˚\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227 \text{ \AA}. Alternatively, using λ=h2mqV=1.23×10−10 m\lambda = \frac{h}{\sqrt{2mqV}} = 1.23 \times 10^{-10} \text{ m}.

Explanation:

The de-Broglie wavelength of a charged particle depends on the accelerating potential VV. For electrons, the simplified expression derived from mm, ee, and hh is very efficient.