The work function for the following metals is given: Na: $2.75\ \text{eV}$; K: $2.30\ \text{eV}$; Mo: $4.17\ \text{eV}$; Ni: $5.15\ \text{eV}$. Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\ \text{Å}$ from a He-Cd laser placed $1\ \text{m}$ away from the photocell? What happens if the laser is brought nearer and placed $50\ \text{cm}$ away?

The work function for the following metals is given: Na: 2.75eV; K: 2.30eV; Mo: 4.17eV; Ni: 5.15eV. Which of t

Additional Exercises · 11.29

2.75\ \text{eV}

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Concept understanding — Photoelectric Effect

The Photoelectric Effect: When Light Knocks Electrons Loose

Imagine you're throwing tennis balls at a wall covered in loose pebbles. If you throw hard enough, a pebble might get knocked off. That's the basic picture — but the photoelectric effect is the quantum version of this, and it completely shattered classical physics.

The Intuition

Light is made of tiny packets of energy called photons. Each photon carries a specific amount of energy, determined by its colour (frequency). When a photon hits a metal surface, it can transfer its energy to an electron inside the metal. If that energy is enough, the electron breaks free and flies out.

Think of electrons in a metal like people in a room with a high window. To escape, they need enough energy to reach the window sill. A photon is like a boost — but only if it gives enough energy in one shot. No amount of weak boosts (dim light) will work if each individual boost is too small.

The Precise Statement

E_{\text{photon}} = hf = \phi + K_{\text{max}}

Where:

$E_{\text{photon}} = hf$ is the energy of a photon (Planck's constant $h = 6.63 \times 10^{-34} \ \text{J·s}$ , $f$ is frequency)
$\phi$ is the work function — the minimum energy needed to remove an electron from that metal
$K_{\text{max}}$ is the maximum kinetic energy of the ejected electron

What Classical Physics Got Wrong

Before Einstein (1905), physicists thought light was a continuous wave. They expected:

Brighter light → more energy per electron → faster electrons
Any colour would eventually eject electrons if you waited long enough

But experiments showed the opposite:

Observation	Classical Prediction	Actual Result
Effect of intensity	Brighter light → faster electrons	Brighter light → more electrons, same speed
Threshold frequency	None — any light works eventually	Below a certain frequency, no electrons no matter how bright
Time delay	Electrons need time to absorb energy	Electrons appear instantly (within $10^{-9}$ s)

The Key Insight

Einstein said: light behaves like a stream of particles (photons), each with energy $hf$ . One photon interacts with one electron. If $hf < \phi$ , the electron cannot escape — period. If $hf > \phi$ , the excess energy becomes kinetic energy:

K_{\text{max}} = hf - \phi

This is why:

Increasing intensity (more photons) ejects more electrons, but each electron still gets the same energy per photon — so their speed doesn't change.
Below threshold frequency, even a trillion photons per second can't help — each one is too weak individually.

Important

The photoelectric effect proved that light is quantized — it comes in discrete packets. This was the birth of quantum mechanics. Einstein won the 1921 Nobel Prize for this, not for relativity.

A Worked Example

Note

Problem: A metal has work function $\phi = 2.0 \ \text{eV}$ . Light of frequency $f = 6.0 \times 10^{14} \ \text{Hz}$ shines on it. Find the maximum kinetic energy of ejected electrons. ( $h = 4.14 \times 10^{-15} \ \text{eV·s}$ )

Step 1: Photon energy

$E = hf = (4.14 \times 10^{-15})(6.0 \times 10^{14}) = 2.48 \ \text{eV}$ …

Why this formula?

Photoelectric Effect: Why the Key Formulas Hold

The photoelectric effect is a cornerstone of quantum physics. It showed that light behaves as particles (photons) , not just waves. Let's build the reasoning step-by-step.

1. The Core Idea: Energy Conservation

When a photon hits a metal surface, it transfers all its energy to a single electron inside the metal.

The photon's energy is $E = hf$ , where $h$ is Planck's constant and $f$ is the frequency of light.
The electron needs a minimum energy to escape the metal — this is called the work function, $\phi$ .

Why only one electron?

Einstein proposed that light is quantized into discrete packets (photons). A single photon cannot split its energy among multiple electrons — it interacts with one electron at a time.

2. The Photoelectric Equation

If the photon's energy is greater than the work function, the excess energy becomes the electron's kinetic energy after escape:

hf = \phi + K_{\text{max}}

Where:

$hf$ = energy of incident photon
$\phi$ = work function (minimum energy to remove electron)
$K_{\text{max}}$ = maximum kinetic energy of ejected electron

Why "maximum" kinetic energy?

Electrons inside the metal have different binding energies.
Some electrons are near the surface (loosely bound) → get maximum $K$ .
Others are deeper → lose energy in collisions before escaping → lower $K$ .

3. The Stopping Potential Connection

We measure $K_{\text{max}}$ using a stopping potential $V_s$ :

K_{\text{max}} = eV_s

Where $e$ is the electron charge. This is because:

An electric field opposing the electron's motion does work $eV_s$ to stop it.
At the stopping potential, the electron's kinetic energy is exactly balanced by the electric potential energy.

Combining:

hf = \phi + eV_s

This is the Einstein photoelectric equation in its most testable form.

4. Why the Threshold Frequency Exists

From the equation:

hf = \phi + eV_s

If $f$ is too low, $hf < \phi$ . Then:

The photon cannot supply enough energy to overcome the work function.
No electron is ejected, regardless of light intensity.

The threshold frequency $f_0$ is when $K_{\text{max}} = 0$ :

hf_0 = \phi \quad \Rightarrow \quad f_0 = \frac{\phi}{h}

Why intensity doesn't matter for ejection?

Intensity = number of photons per second.
Each photon still has energy $hf$ . If $hf < \phi$ , even a billion photons won't eject an electron — each photon is individually too weak.

5. Why Kinetic Energy Depends on Frequency, Not Intensity

From $K_{\text{max}} = hf - \phi$ :

Frequency $f$ directly determines $K_{\text{max}}$ .
Intensity only affects the number of electrons ejected (more photons → more electrons), not their individual energy.

This was the key experimental contradiction with classical wave theory:

Classical: Higher intensity = bigger wave amplitude = more energy to electrons.
Reality: Higher frequency = more energy per electron; intensity only changes current.

6. Summary of Key Relationships …

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The 3300 Å photon carries about 3.77 eV. Comparing this to each metal's work function shows Na and K (both below 3.77 eV) will emit photoelectrons, while Mo and Ni (both above 3.77 eV) will not — and this stays true even if the laser is brought closer, since intensity cannot compensate for insufficient photon energy.

Step 1 — Energy of each incident photon.

E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{3300\times10^{-10}} = \frac{1.989\times10^{-25}}{3.3\times10^{-7}} \approx 6.03\times10^{-19}\ \text{J}

E \approx \frac{6.03\times10^{-19}}{1.6\times10^{-19}} \approx 3.77\ \text{eV}

Step 2 — Compare against each metal's work function.

Metal	$\phi$ (eV)	$E$ vs $\phi$	Emission?
Na	2.75	$3.77 > 2.75$	Yes
K	2.30	$3.77 > 2.30$	Yes
Mo	4.17	$3.77 < 4.17$	No
Ni	5.15	$3.77 < 5.15$	No

So molybdenum and nickel will not show photoelectric emission with this light, since a single 3300 Å photon simply doesn't carry enough energy to overcome either metal's work function.

Step 3 — Effect of moving the laser to 50 cm (doubling the intensity roughly 4×, by the inverse-square law). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

JKBOSE Class 12 Annual Regular Examination 2023Set ANNUAL1 mark
$Q. What is the effect of decrease in wavelength of incident light on the velocity of photoelectrons ?$
›Reveal solutionSolution

A shorter wavelength means a higher-energy photon, so by Einstein's photoelectric equation the maximum kinetic energy - and hence the maximum velocity - of the emitted photoelectrons increases.

By Einstein's photoelectric equation, when light of frequency nu (wavelength lambda, with nu = c/lambda) falls on a metal surface of work function phi0, the maximum kinetic energy of an emitted photoelectron is

KE_max = hnu - phi0 = (hc/lambda) - phi0 = (1/2) m v_max^2.

Here h is Planck's constant, c the speed of light, m the electron mass. If the wavelength lambda of the incident light is decreased (keeping the metal, and hence phi0, fixed), the photon energy hc/lambda increases. Since phi0 is a fixed property of the metal, KE_max = (hc/lambda) - phi0 increases as lambda decreases.

Because KE_max = (1/2) m v_max^2, an increase in KE_max means the maximum velocity v_max of the emitted photoelectrons also increases:

v_max = sqrt[2(hc/lambda - phi0)/m].

…

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JKBOSE Class 12 Annual Regular Examination 2022Set SZ1 mark
$Q. Define threshold frequency.$
›Reveal solutionSolution

Threshold frequency is the minimum light frequency needed to just eject an electron from a metal surface, related to the work function by $W_0 = h\nu_0$ .

When light of a certain frequency falls on a metal surface, photoelectrons are emitted only if the frequency of the incident light is high enough to supply at least the minimum energy needed to free an electron from the metal (the work function $W_0$ ).

The threshold frequency, denoted $\nu_0$ , is defined as the minimum frequency of incident radiation which can just cause photoelectric emission from a given metal surface, with zero kinetic energy given to the emitted electron. For frequencies below $\nu_0$ , no photoelectrons are emitted at all, regardless of how intense (bright) the incident light is — only the number of electrons emitted increases with intensity, not whether emission occurs.

…

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JKBOSE Class 12 Annual Regular Examination 2021Set SZ1 mark
$Q. Define Work Function.$
›Reveal solutionSolution

Work function is the minimum binding energy that must be supplied to free the most loosely bound electron from a metal surface.

In a metal, free electrons are held inside by the attractive forces of positive ions; energy is needed to pull an electron out through the surface.

The work function $\phi_0$ of a metal is defined as the minimum energy required to just liberate an electron from the metal surface, so that it can escape with zero kinetic energy.

It is related to the threshold frequency $\nu_0$ (the minimum frequency of incident light that can cause photoemission) by

$\phi_0 = h u_0 = \frac{hc}{\lambda_0}$

where $h$ is Planck's constant and $\lambda_0$ is the threshold wavelength.

…

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JKBOSE Class 12 Annual Regular Examination 2018Set ANNUAL1 markMCQ
$Q. Work function is measured in :$
(a) Electrovolt
(b) Volt
(c) Joul/sec
(d) Watt
›Reveal solutionSolution

Work function is an energy, and by convention is expressed in electronvolts because joules would be inconveniently small numbers.

Work function ( $\phi_0$ ) is the minimum energy required to just liberate an electron from the surface of a metal (with zero kinetic energy). Being an energy, its SI unit is the joule, but because typical work function values are extremely small in joules (of order $10^{-19}$ J), it is conventionally expressed in electronvolts (eV), where $1\ \text{eV} = 1.6\times10^{-19}$ J — a far more convenient scale (typ …

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