The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold fre

Exercises · 2.51

Q. The work function for caesium atom is 1.9 eV. Calculate

(a) the threshold wavelength and

(b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

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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 sets the minimum photon energy needed to eject an electron. For caesium ( $\phi = 1.9 \text{ eV}$ ), the threshold wavelength is $653 \text{ nm}$ and threshold frequency is $4.59 \times 10^{14} \text{ Hz}$ . When irradiated with $500 \text{ nm}$ light, photoelectrons emerge with kinetic energy $0.58 \text{ eV}$ and velocity $4.52 \times 10^5 \text{ m/s}$ .

The work function $\phi$ represents the minimum energy required to liberate an electron from the metal surface. At threshold, the incident photon carries exactly this energy—no more, no less—so the electron escapes with zero kinetic energy. Any photon with energy above $\phi$ will eject an electron and the surplus appears as kinetic energy, governed by Einstein's photoelectric equation:

E_{\text{photon}} = \phi + KE_{\text{max}}

Since photon energy relates to wavelength by $E = \frac{hc}{\lambda}$ and to frequency by $E = h\nu$ , we can find both threshold quantities and then analyze what happens when the actual wavelength is shorter (higher energy) than threshold.

Part (a): Threshold Wavelength

Set up the threshold condition At threshold, the photon energy equals the work function:

\frac{hc}{\lambda_0} = \phi

Solve for $\lambda_0$

\lambda_0 = \frac{hc}{\phi}

Substitute values Using $h = 6.626 \times 10^{-34} \text{ J·s}$ , $c = 3 \times 10^8 \text{ m/s}$ , and $\phi = 1.9 \text{ eV} = 1.9 \times 1.6 \times 10^{-19} \text{ J} = 3.04 \times 10^{-19} \text{ J}$ :

\lambda_0 = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.04 \times 10^{-19}} = \frac{1.9878 \times 10^{-25}}{3.04 \times 10^{-19}} = 6.54 \times 10^{-7} \text{ m}

$\lambda_0 = 654 \text{ nm}$ (or $653 \text{ nm}$ with more precise constants)

Tip

A quick shortcut: $\lambda_0 (\text{nm}) \approx \frac{1240}{\phi (\text{eV})}$ . Here, $\frac{1240}{1.9} \approx 653 \text{ nm}$ .

Part (b): Threshold Frequency

Relate frequency to work function At threshold:

h\nu_0 = \phi

Solve for $\nu_0$

\nu_0 = \frac{\phi}{h} = \frac{3.04 \times 10^{-19}}{6.626 \times 10^{-34}} = 4.59 \times 10^{14} \text{ Hz}

Alternatively, using $c = \lambda_0 \nu_0$ :

\nu_0 = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{6.54 \times 10^{-7}} = 4.59 \times 10^{14} \text{ Hz}

Kinetic Energy with $\lambda = 500 \text{ nm}$

Calculate the incident photon energy

E_{\text{photon}} = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = \frac{1.9878 \times 10^{-25}}{5 \times 10^{-7}} = 3.976 \times 10^{-19} \text{ J}

Converting to eV:

E_{\text{photon}} = \frac{3.976 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.485 \text{ eV} \approx 2.48 \text{ eV}

Apply Einstein's photoelectric equation …

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Q.The work function for caesium atom is 1.9 eV. Calculate

Part (a): Threshold Wavelength

Part (b): Threshold Frequency

Kinetic Energy with λ=500 nm\lambda = 500 \text{ nm}λ=500 nm

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Kinetic Energy with $\lambda = 500 \text{ nm}$

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