Ultraviolet light of wavelength $2271\ \text{Å}$ from a $100\ \text{W}$ mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is $-1.3\ \text{V}$, estimate the work function of the metal. How would the photo-cell respond to a high intensity ($\sim 10^{5}\ \text{W m}^{-2}$) red light of wavelength $6328\ \text{Å}$ produced by a He-Ne laser?

Ultraviolet light of wavelength 2271Å from a 100W mercury source irradiates a photo-cell made of molybdenum me

Additional Exercises · 11.26

2271\ \text{Å}

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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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From Einstein's equation $KE_{\max}=hc/\lambda - \phi$ with the given stopping voltage, $\phi \approx 4.17$ eV. The red laser's photon energy (~1.96 eV) is below this work function, so no photoelectrons are emitted from it regardless of how intense the beam is made — only photon energy (frequency), not intensity, determines whether emission occurs at all.

Step 1 — Energy of the UV photon.

E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{2271\times10^{-10}} = \frac{1.989\times10^{-25}}{2.271\times10^{-7}} \approx 8.76\times10^{-19}\ \text{J}

E \approx \frac{8.76\times10^{-19}}{1.6\times10^{-19}} \approx 5.47\ \text{eV}

Step 2 — Work function from Einstein's photoelectric equation.

The magnitude of the stopping potential gives the maximum kinetic energy directly: $KE_{\max} = eV_0 = 1.3\ \text{eV}$ .

\phi = E - KE_{\max} = 5.47 - 1.3 \approx 4.17\ \text{eV}

(This matches molybdenum's known work function well.)

Step 3 — Would the red He-Ne laser light cause photoemission?

Photon energy of the red light ( $\lambda=6328\ \text{Å}$ ):

$E_{\text{red}} = \frac{hc}{\lambda} = \frac{1.989\times10^{-25}}{6.328\times10^{-7}} \approx 3.14\times10^{-19}\ \text{J} \approx 1.96\ \text{eV}$ …

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