The photoelectric cut-off voltage in a certain experiment is $1.5\ \text{V}$. What is the maximum kinetic energy of photoelectrons emitted?

> [!TLDR] > The maximum kinetic energy of photoelectrons equals the stopping potential times the electron charge. Here, $K_{\text{max}} = 1.5\ \text{eV}$ or $2.4 \times 10^{-19}\ \text{J}$. The photoelectric effect is one of those rare experiments where a single measurement — the cut-off (or stopping) voltage — directly gives you the maximum kinetic energy of the emitted electrons. No need to know the work function or the incident light frequency. That’s the beauty of it. **Why does this work?** When you apply a reverse voltage between the emitter and collector, you create an electric field that opposes the motion of photoelectrons. The most energetic electrons — those with maximum kinetic energy — are the hardest to stop. The cut-off voltage $V_0$ is exactly the voltage needed to bring these fastest electrons to rest just as they reach the collector. At that point, the electrical potential energy gained ($eV_0$) equals the kinetic energy lost. So the relation is direct: > [!FORMULA] > $$K_{\text{max}} = e V_0$$ where $e = 1.6 \times 10^{-19}\ \text{C}$ is the elementary charge. Now let’s apply it. 1. **Identify the given data.** The cut-off voltage is $V_0 = 1.5\ \text{V}$. 2. **Write the formula.** $$K_{\text{max}} = e V_0$$ 3. **Compute in electronvolts (eV).** Since $e \times 1\ \text{V} = 1\ \text{eV}$, the answer in eV is simply the numerical value of $V_0$: $$K_{\text{max}} = 1.5\ \text{eV}$$ 4. **Convert to joules (SI unit).** Multiply by $e$: $$K_{\text{max}} = (1.6 \times 10^{-19}\ \text{C}) \times (1.5\ \text{V}) = 2.4 \times 10^{-19}\ \text{J}$$ > [!WARNING] > A common mistake is to forget that the cut-off voltage *is* the stopping potential — it’s already the voltage that stops the fastest electrons. Do not multiply by anything extra like the work function or frequency. The relation $K_{\text{max}} = eV_0$ is complete. > [!TIP] > In photoelectric problems, always check whether the answer is expected in eV or joule

The photoelectric cut-off voltage in a certain experiment is 1.5V. What is the maximum kinetic energy of photo

Exercises · 11.3

1.5\ \text{V}

Punjab PsebTextbookSubjective· 2mImportance★★★★★est

7% · 6/83 Questions

✓ Free question

Concept understanding — Maximum Kinetic Energy

Maximum Kinetic Energy – From Intuition to Precision

Kinetic energy is the energy of motion: the faster something moves, the more kinetic energy it has. In many physical situations there is a maximum possible kinetic energy a particle can reach — set either by energy conservation or by an external energy constraint.

The Precise Statement

K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2

Where:

$K_{\text{max}}$ = maximum kinetic energy (in joules)
$m$ = mass of the object (in kg)
$v_{\text{max}}$ = maximum speed reached (in m/s)

This formula alone doesn't tell you why there's a maximum — the physics lies in energy conservation or in an external constraint that limits the speed.

Where Does the Maximum Come From?

1. Energy conservation (no friction)

In a closed system, total mechanical energy $E = K + U$ is constant, so

K_{\text{max}} = E - U_{\text{min}}

The maximum kinetic energy occurs when the potential energy $U$ is at its minimum — for example, a falling object is fastest (and $U$ smallest) just before it lands.

2. External constraints (e.g., the photoelectric effect)

In modern physics, electrons in a metal absorb light energy. Each photon delivers a fixed energy $hf$ . The electron must spend part of that energy escaping the metal (the work function $\phi$ ); the rest becomes kinetic energy:

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

Here the maximum is set entirely by the photon energy — no matter how intense the light, no single electron can gain more kinetic energy than this.

A Common Mistake

Watch out

Students often think "maximum kinetic energy" means the fastest speed possible in the universe. It doesn't. The "maximum" is relative to the given system — the highest value under the stated conditions (height, spring compression, photon energy, and so on), not a universal speed limit.

Important

Maximum kinetic energy is the kinetic energy at the point of greatest speed in a given situation. Find it by energy conservation ( $K_{\text{max}} = E - U_{\text{min}}$ ) or by subtracting any "escape" energy from the input energy ( $K_{\text{max}} = \text{input} - \text{threshold}$ ). Always identify what limits the speed — that's where the maximum comes from.

Maximum kinetic energy calculations, especially via the photoelectric equation, are a staple of the CBSE Class 12 Physics chapter on Dual Nature of Radiation and Matter, and are a high-frequency topic in "photoelectric effect important questions" for JEE Main and NEET. Because this idea also connects to general energy-conservation problems in mechanics, it is worth mastering both as a standalone NCERT-aligned concept and as a recurring numerical type across competitive physics papers.

Why this formula?

Maximum Kinetic Energy — Why the Formula Holds

The idea of "maximum kinetic energy" appears in two very different contexts in your syllabus: photoelectric effect (modern physics) and simple harmonic motion (oscillations). I'll cover both, because the why is different in each case.

1. In the Photoelectric Effect

The formula you must know:

K_{\text{max}} = h\nu - \phi

where $h$ is Planck's constant, $\nu$ is the frequency of incident light, and $\phi$ is the work function of the metal.

Why this formula? It comes from Einstein's photon model and energy conservation.

A single photon carries energy $E = h\nu$ . When it strikes a metal surface, it can transfer all of its energy to one electron. That electron must first overcome the binding force holding it in the metal — the minimum energy needed for this is the work function $\phi$ . Any leftover energy becomes the electron's kinetic energy after it escapes.

So:

Photon energy = Energy to escape + Kinetic energy of ejected electron

h\nu = \phi + K

If the electron just barely escapes (with zero kinetic energy), the photon frequency is the threshold frequency $\nu_0$ , where $h\nu_0 = \phi$ .

For a higher frequency, the maximum kinetic energy an ejected electron can have is when it absorbs the photon's full energy and loses nothing to collisions inside the metal. That gives:

K_{\text{max}} = h\nu - \phi

Watch out

$K_{\text{max}}$ does not depend on light intensity. Intensity only increases the number of electrons, not their maximum energy. This was the key puzzle that classical physics couldn't explain.

2. In Simple Harmonic Motion (SHM)

For a particle executing SHM, the maximum kinetic energy is:

K_{\text{max}} = \frac{1}{2} m \omega^2 A^2

where $m$ is mass, $\omega$ is angular frequency, and $A$ is amplitude.

Why this formula? It follows directly from the velocity equation.

In SHM, displacement is $x = A \sin(\omega t + \phi)$ . Differentiating gives velocity:

v = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)

The velocity is maximum when $\cos(\omega t + \phi) = \pm 1$ , i.e., at the equilibrium position ( $x = 0$ ):

v_{\text{max}} = A\omega

Kinetic energy is $K = \frac{1}{2} m v^2$ , so:

K_{\text{max}} = \frac{1}{2} m (A\omega)^2 = \frac{1}{2} m \omega^2 A^2

Note

At the extreme positions ( $x = \pm A$ ), velocity is zero, so $K = 0$ . All the energy is potential. At equilibrium, all energy is kinetic. The total mechanical energy $E = \frac{1}{2} m \omega^2 A^2$ is constant and equals $K_{\text{max}}$ .

Quick Comparison

Context	Formula for $K_{\text{max}}$	Key Insight
Photoelectric effect	$h\nu - \phi$	Energy conservation per photon; independent of intensity
SHM	$\frac{1}{2} m \omega^2 A^2$	Velocity is maximum at equilibrium; $v_{\text{max}} = A\omega$

Important

In photoelectric problems, $K_{\text{max}}$ is often found by measuring the stopping potential $V_0$ : $K_{\text{max}} = eV_0$ . This is a direct experimental link — the stopping potential just balances the maximum kinetic energy of the fastest electrons.

←Previous question11.2 — The work function of caesium metal is 2.14eV. When light of frequency …→Next question11.4 — Monochromatic light of wavelength 632.8nm is produced by a helium-neon…

Q.The photoelectric cut-off voltage in a certain experiment is 1.5 V1.5\ \text{V}1.5 V. What is the maximum kinetic energy of photoelectrons emitted?