Threshold frequency, ν0 is the minimum frequency which a photon must possess to eject an electron from a metal

NCERT Exemplar · Q52

Q. Threshold frequency, ν0 is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency 1.0\times10^15 s^-1 was allowed to hit a metal surface, an electron having 1.988 \times 10^-19 J of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.

Tripura TbseLong· 3mImportance★★★★★est

98% · 137/140 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

The photoelectric effect equation $KE = h\nu - h\nu_0$ gives the threshold frequency. For the given data, $\nu_0 = 7.0 \times 10^{14} \text{ s}^{-1}$ . A 600 nm photon has frequency $5.0 \times 10^{14} \text{ s}^{-1}$ , which is below threshold, so no emission occurs.

The photoelectric effect is a clean demonstration of quantum energy transfer. A photon delivers its entire energy $h\nu$ 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 $W = h\nu_0$ . The leftover energy appears as the electron's kinetic energy. So the equation is simply energy conservation at the quantum level:

KE = h\nu - h\nu_0

We know $h = 6.626 \times 10^{-34} \text{ J·s}$ , a universal constant. The threshold frequency $\nu_0$ is what we need — the minimum frequency for which the photon energy just equals the work function, giving zero kinetic energy.

Write the photoelectric equation and substitute known values. The kinetic energy is $1.988 \times 10^{-19} \text{ J}$ , the incident frequency is $\nu = 1.0 \times 10^{15} \text{ s}^{-1}$ , and Planck's constant $h = 6.626 \times 10^{-34} \text{ J·s}$ . So:

1.988 \times 10^{-19} = (6.626 \times 10^{-34})(1.0 \times 10^{15}) - (6.626 \times 10^{-34})\nu_0

Calculate the incident photon energy.

h\nu = (6.626 \times 10^{-34})(1.0 \times 10^{15}) = 6.626 \times 10^{-19} \text{ J}

Solve for $\nu_0$ . Rearranging the equation:

h\nu_0 = h\nu - KE = 6.626 \times 10^{-19} - 1.988 \times 10^{-19} = 4.638 \times 10^{-19} \text{ J}

Then:

\nu_0 = \frac{4.638 \times 10^{-19}}{6.626 \times 10^{-34}} = 7.0 \times 10^{14} \text{ s}^{-1}

Tip

Notice that $1.988 \times 10^{-19}$ is very close to $0.3 \times 6.626 \times 10^{-19}$ , so the threshold energy is $0.7 \times h\nu$ — a quick mental check that $\nu_0 = 0.7 \times 10^{15} = 7 \times 10^{14}$ .

Now check the 600 nm photon. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ51 — What is photoelectric effect? State the result of photoelectric effect…→Next questionQ53 — When an electric discharge is passed through hydrogen gas, the hydroge…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days