Light of wavelength $488\ \text{nm}$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38\ \text{V}$. Find the work function of the material from which the emitter is made.

Light of wavelength 488nm is produced by an argon laser which is used in the photoelectric effect. When light

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488\ \text{nm}

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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. …

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 ( $x = 0$ ):

$v_{\text{max}} = A\omega$ …

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Einstein's equation gives $\phi = \dfrac{hc}{\lambda}-eV_0 = 2.54\ \text{eV}-0.38\ \text{eV} = 2.16\ \text{eV}$ .

Einstein's photoelectric equation

When a photon of energy $hc/\lambda$ frees an electron, part goes to overcome the work function $\phi$ and the rest becomes kinetic energy. The stopping potential $V_0$ just halts the fastest electrons, so $K_{\max}=eV_0$ :

\frac{hc}{\lambda} = \phi + eV_0.

Step 1 — Photon energy of the 488 nm line

Using $hc \approx 1240\ \text{eV·nm}$ ,

E = \frac{hc}{\lambda} = \frac{1240\ \text{eV·nm}}{488\ \text{nm}} = 2.54\ \text{eV}.

Step 2 — Maximum kinetic energy

The cut-off (stopping) potential is $V_0 = 0.38$ V, so

$K_{\max} = eV_0 = 0.38\ \text{eV}.$ …

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Common Mistakes & How to Avoid Them

Mistake 1: Forgetting to convert wavelength to metres

Students often plug $488$ directly into $E = \frac{hc}{\lambda}$ without converting nanometres to metres. Since $h$ is in J·s and $c$ in m/s, $\lambda$ must be in metres.

How to avoid: Always write the conversion explicitly:

$\lambda = 488\ \text{nm} = 488 \times 10^{-9}\ \text{m} = 4.88 \times 10^{-7}\ \text{m}$

Do this as your first step, before touching any formula.

Mistake 2: Using the wrong value of $h$ or $c$

Some students use $h = 6.63 \times 10^{-34}\ \text{J·s}$ but then forget to use $c = 3 \times 10^8\ \text{m/s}$ , or they mix up $h$ and $\hbar$ .

How to avoid: Write the constants clearly at the top of your solution:

$h = 6.63 \times 10^{-34}\ \text{J·s}$ , $c = 3 \times 10^8\ \text{m/s}$ , $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$

Keep them visible throughout the calculation.

Mistake 3: Confusing stopping potential with kinetic energy

The stopping potential $V_0$ is not the kinetic energy — it is the potential difference that stops the most energetic electrons. The maximum kinetic energy is $K_{\text{max}} = eV_0$ , where $e$ is the electron charge.

How to avoid: Remember the photoelectric equation in its two equivalent forms:

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

The second form directly uses the stopping potential. Never write $K_{\text{max}} = V_0$ — that is dimensionally wrong.

Mistake 4: Forgetting to convert electron-volt answers to joules (or vice versa)

The work function is often asked in eV, but $h$ and $c$ give energy in joules. Students either forget to convert at all, or use the wrong conversion factor.

How to avoid: Calculate the photon energy in joules first, then convert to eV at the end if needed.

$E_{\text{photon}} = \frac{hc}{\lambda}$ gives joules.

$1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$ , so divide by this to get eV.

Mistake 5: Sign errors in the photoelectric equation

Students sometimes write $hf + \phi = eV_0$ or $hf = \phi - eV_0$ , both of which are incorrect.

How to avoid: The photon energy is split into two parts: work function (to free the electron) and kinetic energy (what remains). So:

Important

$hf = \phi + eV_0$

$\phi = hf - eV_0$ …

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Einstein's photoelectric equation

Step 1 — Photon energy of the 488 nm line

Step 2 — Maximum kinetic energy

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