Photoelectric Effect — Definition, Experiment, Photocurrent Variation, Practice Problems & FAQs

Photoelectric Effect — Definition, Experimental Study, Photocurrent Variation, Practice Problems & FAQs

Written By Mohan Dangi (Gold Medalist)

Table of contents

• Photoelectric effect — definition
• Experimental setup (photoelectric tube)
• Variation of photocurrent with: potential, intensity & frequency
• Einstein's photoelectric equation
• Practice problems (with solutions)
• FAQs

Photoelectric effect — definition

The photoelectric effect is the emission of electrons (called photoelectrons) from the surface of a metal when light (electromagnetic radiation) of sufficiently high frequency falls on it. The resulting current produced when these electrons are collected is called the photocurrent.

Historically: Heinrich Hertz observed related phenomena while investigating sparks; thermionic emission was known (electrons emitted on heating). The photoelectric effect showed that light must sometimes behave like particles (photons) since the classical wave picture couldn't explain key observations (threshold frequency, instantaneous emission, energy independent of intensity).

Experimental setup (photoelectric tube)

Figure: Typical photoelectric tube with photosensitive cathode (C) and collector anode (A). Light enters through a quartz window.

The common setup includes:

• An evacuated glass tube containing a photosensitive material as the cathode (C) and a collector as the anode (A).
• Light of known frequency & intensity is made incident on the cathode via a quartz window.
• A variable voltage (bias) is applied between anode and cathode, and a microammeter measures the photocurrent.

Variation of photocurrent with different parameters

Effect of intensity (at fixed frequency)

If the incident frequency v is fixed above the threshold frequency v₀, increasing the intensity (number of photons per second) increases the number of emitted electrons per second — therefore the saturation photocurrent increases proportionally. However, the maximum kinetic energy of emitted electrons does not change with intensity.

Effect of anode-cathode potential (bias)

With the anode positive (accelerating), more emitted electrons are collected and the current rises until it saturates (all emitted electrons are collected). At zero bias, some current exists because electrons have kinetic energy. When the anode is made negative (retarding potential), fewer electrons reach the anode, and at a certain negative potential (the stopping potential or V₀) the photocurrent drops to zero — the most energetic electrons cannot reach the anode.

Effect of frequency (at fixed intensity)

Increasing the frequency (for v > v₀) increases the maximum kinetic energy of photoelectrons, so the stopping potential becomes more negative (|V₀| increases). The saturation current (number of electrons emitted) remains essentially unchanged if intensity is constant.

Summary of experimental signatures that contradict classical wave theory

• A minimum (threshold) frequency v₀ is required — below v₀ no electrons are emitted regardless of intensity.
• Photoemission is essentially instantaneous (no measurable time lag) above threshold.
• Maximum kinetic energy of electrons depends on frequency, not intensity.
• Saturation current depends on intensity (number of photons), not on frequency (above v₀).

Einstein’s photoelectric equation

Einstein explained the effect by proposing that light consists of quanta (photons) each with energy \(E = h\nu\) (where \(h\) is Planck’s constant, \(\nu\) the frequency). When a photon is absorbed by an electron, the photon energy is used to overcome the metal’s work function \(\phi\) and the rest becomes the kinetic energy of the emitted electron:

\[ K_{\text{max}} = h\nu - \phi \] where \(\phi = h\nu_0\) (work function) and \(\nu_0\) is the threshold frequency.

The stopping potential \(V_0\) relates to \(K_{\text{max}}\) by \(eV_0 = K_{\text{max}}\) (where \(e\) is the elementary charge). Thus,

\[ eV_0 = h\nu - \phi \] or rearranged, \[ V_0 = \frac{h}{e}\nu - \frac{\phi}{e}. \]

Practice problems (with solutions)

Problem 1 (which metals show photoemission?)

The work functions \(\phi\) of some metals are listed below (in eV):

• Li: 2.4 eV
• Cu: 4.8 eV
• Ag: 4.3 eV
• W: 4.75 eV

Which of these metals will show photoelectric emission when illuminated by light of wavelength \(300\ \text{nm}\)? (Use \(hc = 1240\ \text{eV·nm}\))

Photon energy \(E = \dfrac{hc}{\lambda} = \dfrac{1240\ \text{eV·nm}}{300\ \text{nm}} = 4.133\ \text{eV}.\) Photoemission occurs if \(E > \phi\).

• Li: \(4.133 > 2.4\) → emits
• Cu: \(4.133 < 4.8\) → no
• Ag: \(4.133 < 4.3\) → no
• W: \(4.133 < 4.75\) → no

Answer: Only Li shows photoelectric emission at 300 nm.

Problem 2 (find work function from speeds)

A metal is illuminated by two wavelengths 250 nm and 300 nm. The maximum speeds of ejected electrons are \(v_1\) and \(v_2\) respectively, and \(v_1 : v_2 = 2 : 1\). Find the work function \(\phi\) of the metal. Use \(hc = 1240\ \text{eV·nm}\).

Let \(K_1 = \tfrac{1}{2}m v_1^2\), \(K_2 = \tfrac{1}{2}m v_2^2\). Given \(v_1 = 2 v_2\) ⇒ \(K_1 = \tfrac{1}{2} m (2v_2)^2 = 4 \cdot \tfrac{1}{2} m v_2^2 = 4K_2\). From Einstein’s equation: \[ K_1 = h\nu_1 - \phi,\qquad K_2 = h\nu_2 - \phi. \] Since \(K_1 = 4K_2\): \[ h\nu_1 - \phi = 4(h\nu_2 - \phi). \] Rearranged: \[ h\nu_1 - \phi = 4h\nu_2 - 4\phi \Rightarrow 3\phi = 4h\nu_2 - h\nu_1. \] Substitute \(\nu = c/\lambda\) and use \(hc = 1240\ \text{eV·nm}\): \[ 3\phi = 4\frac{1240}{300} - \frac{1240}{250}\ \text{eV}. \] Compute: \[ 4\frac{1240}{300} = \frac{4960}{300} = 16.533\ \text{eV},\quad \frac{1240}{250} = 4.96\ \text{eV}. \] So \[ 3\phi = 16.533 - 4.96 = 11.573\ \text{eV} \Rightarrow \phi = \frac{11.573}{3} = 3.8577\ \text{eV}. \] Answer: \(\phi \approx 3.86\ \text{eV}.\)

Problem 3 (threshold frequency & no emission)

A metal with threshold frequency \( \nu_0 = 1.0 \times 10^{14}\ \text{Hz}\) is struck by light of frequency \(6.0 \times 10^{13}\ \text{Hz}\). Will electrons be emitted?

Since \(\nu_{\text{incident}} < \nu_0\), the photon energy is less than the work function, so no photoelectrons will be emitted.

Problem 4 (from a \(V_0\) vs \(\nu\) graph)

If a graph of stopping potential \(V_0\) (V) versus frequency \(\nu\) (Hz) intercepts the frequency axis at \(\nu_0 = 4.0 \times 10^{15}\ \text{Hz}\), find the work function \(\phi\). Take \(h = 6.6 \times 10^{-34}\ \text{J·s}\).

\[ \phi = h\nu_0 = (6.6\times10^{-34}\ \text{J·s})(4.0\times10^{15}\ \text{s}^{-1}) = 26.4\times10^{-19}\ \text{J}. \] Convert to eV: \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\). \[ \phi = \frac{26.4\times10^{-19}}{1.6\times10^{-19}} = 16.5\ \text{eV}. \] Answer: \(\phi = 16.5\ \text{eV}.\)

FAQs

Why do photoelectrons not all come out with the same energy for a monochromatic light source?

Although every photon has the same energy \(h\nu\), electrons in the metal occupy different energy states (different binding depths). When a photon ejects an electron, some of the energy first overcomes the local binding (work function for that electron); the remainder is kinetic energy. Therefore the emitted electrons show a range of kinetic energies up to the maximum \(K_{\text{max}} = h\nu - \phi\).

Why can't the classical wave theory of light explain the photoelectric effect?

Wave theory predicts increasing intensity should increase electron energy and also predicts a time lag for low intensity light. Experiments show emission is instantaneous (above threshold) and kinetic energy depends on frequency not intensity — facts inconsistent with a pure wave picture but explained by photons (quantized energy packets).

What are practical applications of the photoelectric effect?

Photocells/phototubes, light meters, photo-detectors used in cameras, automatic doors, solar cells (different mechanism but related in converting light to electrical energy), and historically in development of quantum mechanics.

How can photocurrent be increased in an experiment?

Increase the intensity (more photons per second) or increase the effective collecting area or improve the cathode material’s photoemissive efficiency. Bringing the light source closer (higher intensity) or using a higher-power lamp increases saturation photocurrent. Increasing frequency (while keeping intensity constant) does not increase photocurrent but increases electron energy.

What is stopping potential and how is it measured?

Stopping potential \(V_0\) is the (negative) potential applied to the anode relative to cathode that just prevents the most energetic photoelectrons from reaching the anode (photocurrent = 0). It is measured by making the bias more retarding until the photocurrent falls to zero; then \(eV_0 = K_{\text{max}}.\)

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