The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped

Exercises · 2.53

Q. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

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Concept understanding — Photon Energy Calculation

Photon Energy Calculation

The Intuition First

Imagine you're holding a rope tied to a wall. If you flick your wrist once, a single pulse travels down the rope. If you flick faster — more frequently — each pulse carries more energy; the rope vibrates more violently. Light behaves the same way. A photon is the smallest possible "flick" of the electromagnetic field — a single, indivisible packet of light energy.

What determines how much energy that one photon carries? Two things: how fast the wave is oscillating (its frequency), and a universal constant that connects the wave world to the particle world.

The Precise Statement

The energy $E$ of a single photon is directly proportional to its frequency $f$ (or inversely proportional to its wavelength $\lambda$ ). The proportionality constant is Planck's constant $h$ , one of the most fundamental numbers in physics.

E = h f = \frac{h c}{\lambda}

Where:

$E$ = energy of one photon (in joules, J)
$h$ = Planck's constant = $6.626 \times 10^{-34} \ \text{J·s}$
$f$ = frequency of the electromagnetic wave (in hertz, Hz)
$c$ = speed of light = $3.00 \times 10^{8} \ \text{m/s}$
$\lambda$ = wavelength of the light (in metres, m)

Why Two Forms?

The first form $E = hf$ is the most direct: higher frequency means higher energy. The second form $E = hc/\lambda$ is often more practical because wavelength is easier to measure than frequency. Since $c = f\lambda$ , you can always swap between them.

What This Tells You

Blue light (short wavelength, high frequency) has more energy per photon than red light (long wavelength, low frequency).
Gamma rays have enormous photon energies; radio waves have tiny photon energies.
The energy is quantised — you cannot have half a photon. Either the full energy $hf$ is absorbed/emitted, or none at all.

A Worked Example

Question: Calculate the energy of a single photon of violet light with wavelength $400 \ \text{nm}$ .

Step 1: Convert wavelength to metres.

400 \ \text{nm} = 400 \times 10^{-9} \ \text{m} = 4.00 \times 10^{-7} \ \text{m}

Step 2: Use $E = hc/\lambda$ .

E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.00 \times 10^{-7}}

Step 3: Compute.

E = \frac{1.9878 \times 10^{-25}}{4.00 \times 10^{-7}} = 4.97 \times 10^{-19} \ \text{J}

Note

This is an incredibly tiny amount of energy — about $5 \times 10^{-19}$ joules. That's why we often use electronvolts (eV) for photon energies in atomic physics. $1 \ \text{eV} = 1.602 \times 10^{-19} \ \text{J}$ , so this photon has about $3.1 \ \text{eV}$ .

Common Mistake to Avoid …

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The work function is the minimum energy needed to eject an electron. Using the photoelectric equation $eV_0 = h\nu - \phi$ , we find $\phi = 4.48 \, \text{eV}$ for silver.

The photoelectric effect is a clean demonstration of light behaving as particles (photons). When a photon hits a metal surface, it transfers its entire energy to an electron. Part of that energy goes into overcoming the binding force that holds the electron in the metal — that's the work function ( $\phi$ ). The leftover energy appears as the electron's kinetic energy.

The stopping voltage $V_0$ is a clever experimental tool: it's the reverse voltage that just barely stops the most energetic photoelectrons from reaching the other electrode. At that point, the electrical potential energy $eV_0$ exactly equals the maximum kinetic energy of the ejected electrons. So we can write:

eV_0 = h\nu - \phi

where $e = 1.6 \times 10^{-19} \, \text{C}$ , $h = 6.63 \times 10^{-34} \, \text{J·s}$ , and $\nu = c/\lambda$ .

Let's work through the numbers.

Find the photon frequency. The wavelength is $\lambda = 256.7 \, \text{nm} = 256.7 \times 10^{-9} \, \text{m}$ . Using $c = 3 \times 10^8 \, \text{m/s}$ :

\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{256.7 \times 10^{-9}} = 1.168 \times 10^{15} \, \text{Hz}

Calculate the photon energy in joules.

E_{\text{photon}} = h\nu = (6.63 \times 10^{-34})(1.168 \times 10^{15}) = 7.74 \times 10^{-19} \, \text{J}

Convert the stopping voltage to energy. The stopping voltage is $V_0 = 0.35 \, \text{V}$ , so:

eV_0 = (1.6 \times 10^{-19})(0.35) = 5.6 \times 10^{-20} \, \text{J}

Apply the photoelectric equation. $eV_0 = h\nu - \phi \quad \Rightarrow \quad \phi = h\nu - eV_0$ …

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Q.The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

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