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Maximum Kinetic Energy of Ejected Electrons
1.8357
eV
Electron emitted? Yes
Photon energy 4.1357 eV
Threshold frequency f₀ 5.5614 ×10¹⁴ Hz

What is the Photoelectric Effect Calculator?

This tool applies Einstein's photoelectric equation to find the maximum kinetic energy of electrons ejected from a metal surface when light of a given frequency strikes it. It is a universal physics calculator based on fundamental constants, so it applies anywhere.

Photon striking a metal surface and ejecting an electron
Light hitting a metal ejects electrons when each photon's energy exceeds the work function.

How to use it

Enter the frequency of the incoming light in units of \(\times 10^{14}\) Hz (visible light is roughly 4–7.5 \(\times 10^{14}\) Hz) and the work function \(\Phi\) of the metal in electron-volts (eV). The calculator returns the photon energy, the maximum electron kinetic energy, the threshold frequency, and whether emission occurs.

The formula explained

The governing equation is $$KE_{max} = h \cdot f - \Phi$$ where \(h\) is Planck's constant (\(6.626 \times 10^{-34}\ \text{J}\cdot\text{s}\)), \(f\) is the light frequency, and \(\Phi\) is the work function. A photon carries energy \(E = h \cdot f\). If that energy exceeds \(\Phi\), the surplus becomes the electron's kinetic energy; if not, no electron is emitted. The threshold frequency is \(f_0 = \Phi / h\).

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Energy bar diagram showing photon energy split into work function and kinetic energy
Photon energy \(hf\) splits into the work function \(\Phi\) and the electron's kinetic energy.

Worked example

For \(f = 10 \times 10^{14}\) Hz (\(1.0 \times 10^{15}\) Hz) and \(\Phi = 2.3\) eV: photon energy $$= \frac{h \cdot f}{e} = \frac{6.626 \times 10^{-34} \times 1.0 \times 10^{15}}{1.602 \times 10^{-19}} \approx 4.136\ \text{eV}$$ $$KE_{max} = 4.136 - 2.3 \approx 1.836\ \text{eV}$$ Threshold $$f_0 = \frac{2.3 \times 1.602 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 5.56 \times 10^{14}\ \text{Hz}$$ Since the photon exceeds the work function, electrons are emitted.

FAQ

What if the photon energy is below the work function? No electrons are ejected; \(KE_{max}\) is reported as 0 and "Electron emitted?" shows No.

Why are frequencies in \(\times 10^{14}\) Hz? Visible and near-visible light frequencies are around \(10^{14}\) Hz, so this scale keeps the input numbers convenient.

Does intensity matter? Intensity affects the number of electrons, not their maximum kinetic energy — only frequency and work function set \(KE_{max}\).

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