Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency ($\nu_0$) and work function ($W_0$) of the metal.

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 68

Exercises · 2.12

\nu_0

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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 threshold frequency is the minimum frequency needed to just eject an electron with zero kinetic energy. Using the photoelectric equation, $\nu_0 = c / \lambda_0$ gives $\nu_0 = 4.41 \times 10^{14} \text{ Hz}$ , and $W_0 = h \nu_0$ gives $W_0 = 2.92 \times 10^{-19} \text{ J}$ .

Why This Approach Works

The photoelectric effect tells us that light behaves as packets of energy called photons. When a photon strikes a metal surface, its energy is used first to overcome the binding force holding the electron to the metal — this minimum energy needed is the work function $W_0$ . Any extra photon energy appears as the electron's kinetic energy.

The problem says electrons are emitted with zero velocity. That means the photon's energy is exactly enough to free the electron, with nothing left over for motion. So the photon energy equals the work function. The wavelength given (6800 Å) is therefore the threshold wavelength $\lambda_0$ — the longest wavelength that can still cause emission.

E_{\text{photon}} = h\nu = \frac{hc}{\lambda}

Step-by-Step Solution

1. Convert wavelength to meters

The given wavelength is 6800 Å. Since $1 \text{ Å} = 10^{-10} \text{ m}$ :

\lambda_0 = 6800 \times 10^{-10} = 6.8 \times 10^{-7} \text{ m}

2. Calculate threshold frequency $\nu_0$

Frequency and wavelength are related by $c = \nu \lambda$ , where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light.

\nu_0 = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{6.8 \times 10^{-7}}

Doing the division:

\nu_0 = 4.41176... \times 10^{14} \text{ Hz}

Rounding to three significant figures:

\nu_0 = 4.41 \times 10^{14} \text{ Hz}

Tip

Notice that frequency is inversely proportional to wavelength. A longer wavelength means a lower frequency — which is why red light (long $\lambda$ ) often fails to eject electrons while blue light (short $\lambda$ ) succeeds.

3. Calculate work function $W_0$

The work function equals the minimum photon energy needed. Using Planck's constant $h = 6.626 \times 10^{-34} \text{ J·s}$ :

W_0 = h \nu_0 = (6.626 \times 10^{-34}) \times (4.41 \times 10^{14})

Multiplying:

W_0 = 2.922 \times 10^{-19} \text{ J}

Rounded appropriately: …

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Q.Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0\nu_0ν0​) and work function (W0W_0W0​) of the metal.

Why This Approach Works

Step-by-Step Solution

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Q.Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency ( $\nu_0$ ) and work function ( $W_0$ ) of the metal.

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