Photoelectric effect — find threshold frequency and maximum kinetic energy

Question

The work function of a metal surface is 4.0 eV. Light of frequency $2.0 \times 10^{15}$ Hz falls on it. Find: (a) the threshold frequency of the metal, and (b) the maximum kinetic energy of the emitted photoelectrons.

Take $h = 6.63 \times 10^{-34}$ J·s and $1\,\text{eV} = 1.6 \times 10^{-19}$ J.

(CBSE 2024, similar pattern)

Solution — Step by Step

The work function $\phi$ is the minimum energy needed to eject an electron. At the threshold frequency $\nu_0$ , the photon has exactly this much energy:

\phi = h\nu_0

\nu_0 = \frac{\phi}{h} = \frac{4.0 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}

\nu_0 = \frac{6.4 \times 10^{-19}}{6.63 \times 10^{-34}} = \mathbf{9.65 \times 10^{14}\,\text{Hz}}

The incident frequency is $2.0 \times 10^{15}$ Hz, which is greater than $\nu_0 = 9.65 \times 10^{14}$ Hz. So yes, photoelectrons will be emitted. If the incident frequency were below $\nu_0$ , no emission would happen regardless of intensity.

KE_{max} = h\nu - \phi

KE_{max} = (6.63 \times 10^{-34})(2.0 \times 10^{15}) - 4.0 \times 1.6 \times 10^{-19}

KE_{max} = 13.26 \times 10^{-19} - 6.4 \times 10^{-19}

KE_{max} = 6.86 \times 10^{-19}\,\text{J}

Converting to eV:

KE_{max} = \frac{6.86 \times 10^{-19}}{1.6 \times 10^{-19}} = \mathbf{4.29\,\text{eV}}

Why This Works

Einstein’s photoelectric equation is an energy conservation statement. A single photon of energy $h\nu$ hits the metal surface and transfers all its energy to one electron. Part of this energy ( $\phi$ ) goes into overcoming the binding forces holding the electron in the metal. The rest becomes the electron’s kinetic energy.

The “maximum” kinetic energy corresponds to electrons at the surface (minimum binding energy = work function). Electrons deeper inside the metal lose additional energy before escaping, so they emerge with less KE.

For NEET and CBSE, memorise $h\nu = \phi + KE_{max}$ and $eV_0 = KE_{max}$ (where $V_0$ is the stopping potential). These two equations solve 90% of photoelectric effect numericals.

Alternative Method

Work entirely in eV to avoid large numbers. The photon energy in eV:

E = h\nu = \frac{6.63 \times 10^{-34} \times 2.0 \times 10^{15}}{1.6 \times 10^{-19}} = \frac{13.26 \times 10^{-19}}{1.6 \times 10^{-19}} = 8.29\,\text{eV}

Then: $KE_{max} = 8.29 - 4.0 = 4.29\,\text{eV}$

Much cleaner. When $\phi$ is given in eV, convert $h\nu$ to eV first and subtract directly.

Common Mistake

The most frequent error: mixing units. Students compute $h\nu$ in joules and subtract $\phi$ in eV directly, getting a nonsensical answer. Always convert to the same unit system before subtracting.

Another trap: confusing threshold frequency with threshold wavelength. The threshold frequency is $\nu_0 = \phi/h$ , and the corresponding threshold wavelength is $\lambda_0 = c/\nu_0 = hc/\phi$ . Note that $\lambda_0$ is the maximum wavelength (not minimum) that can cause emission, since longer wavelength means lower frequency.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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