Dual Nature of Radiation and Matter: Full Notes

11.1 Photoelectric Effect: Experimental Observations

When light of sufficiently high frequency falls on a metal surface, electrons (photoelectrons) are ejected from it. Careful experiments with this effect turned up several facts that no classical wave picture of light could explain together:

Instantaneous emission: photoelectrons appear (when they appear at all) with no measurable time lag, even for very faint light, not after some delay while energy “builds up”.
Existence of a threshold frequency: for each metal there is a minimum frequency $f_0$ below which no photoelectrons are emitted at all, regardless of how intense the incident light is.
Stopping potential is independent of intensity: the retarding voltage needed to stop even the fastest photoelectrons depends only on the light’s frequency, not on its intensity.
Saturation current is proportional to intensity: above threshold, the maximum photocurrent (all emitted electrons collected) rises in direct proportion to the light’s intensity, at fixed frequency.

11.2 Einstein’s Photoelectric Equation

Einstein explained every one of these observations at once by proposing that light itself is quantized: it arrives as discrete packets (photons), each of energy depending only on frequency, never on intensity:

Photon energy

E = hf

An electron absorbs essentially one photon at a time. Some minimum energy, the work function $\varphi$ , is needed just to free the electron from the metal’s surface; whatever energy is left over appears as the electron’s kinetic energy. The fastest-emitted electrons (those that needed the least additional binding energy) define the maximum kinetic energy:

Einstein's photoelectric equation

hf = \varphi + KE_{\max}

This single equation explains the puzzling facts above: emission is instantaneous because a photon either carries enough energy to free an electron right away or it doesn’t, there is no accumulation. A threshold frequency exists because $KE_{\max}=hf-\varphi$ would otherwise be negative, which is unphysical. And since intensity only changes how many photons arrive per second (not each photon’s own energy), raising intensity raises the number of photoelectrons per second (so the saturation current) but leaves $KE_{\max}$ (and hence the stopping potential) completely unchanged.

Worked example

Maximum kinetic energy from Einstein's equation

A metal has work function 2.14 eV. Light of frequency 1.0×10¹⁵ Hz falls on it. Find the maximum kinetic energy of the emitted photoelectrons. (h = 4.1357×10⁻¹⁵ eV·s)

hf = (4.1357\times10^{-15})(1.0\times10^{15}) \approx 4.136\text{ eV}

KE_{\max} = hf-\varphi = 4.136 - 2.14 = \textcolor{#e08a1e}{1.996\text{ eV}}\ (\approx 2.00\text{ eV})

11.2.1 Photon momentum

A photon carries not just energy but momentum too, even though it has no rest mass. For a massless particle, the relativistic energy-momentum relation reduces to $E=pc$ , so combining this with the photon energy $E=hf$ above, and $f=c/\lambda$ :

p = \dfrac{E}{c} = \dfrac{hf}{c} = \dfrac{h(c/\lambda)}{c}

Photon momentum

\textcolor{#e08a1e}{p} = \dfrac{h}{\lambda} = \dfrac{E}{c}

This is precisely the relation de Broglie later turns around to apply to matter (§11.5): light itself already obeys $p=h/\lambda$ , and de Broglie’s leap was to propose that any particle, not just a photon, carries a wavelength given by the same formula, run in reverse.

Worked example

Photon momentum, two ways

Find the momentum of a photon of red light, $\lambda=600\text{ nm}$ , first from its energy and then directly from its wavelength, and confirm the two agree.

f = \dfrac{c}{\lambda} = \dfrac{3.0\times10^8}{600\times10^{-9}} = 5.0\times10^{14}\text{ Hz}, \qquad E = hf = (6.626\times10^{-34})(5.0\times10^{14}) \approx 3.31\times10^{-19}\text{ J}

p = \dfrac{E}{c} = \dfrac{3.31\times10^{-19}}{3.0\times10^8} \approx \textcolor{#e08a1e}{1.10\times10^{-27}\text{ kg·m/s}}

p = \dfrac{h}{\lambda} = \dfrac{6.626\times10^{-34}}{600\times10^{-9}} \approx \textcolor{#e08a1e}{1.10\times10^{-27}\text{ kg·m/s}}

Both routes agree exactly, as they must: $p=E/c$ and $p=h/\lambda$ are the same statement written two ways. This momentum is minuscule, millions of times smaller than the electron momentum in the worked example of §11.5, which is why light pressure is so hard to notice in everyday life.

11.3 Work Function & Threshold Frequency

The work function $\varphi$ is the minimum energy needed to pull an electron out of a given metal’s surface, a property of the metal (and its surface condition), typically a few eV. Setting $KE_{\max}=0$ in Einstein’s equation (electrons just barely freed, with nothing left over) defines the threshold frequency:

Threshold frequency

f_0 = \dfrac{\varphi}{h}

Below $f_0$ , $hf<\varphi$ for every photon, so no single photon ever carries enough energy to free an electron, and since electrons cannot pool energy across multiple photons, emission is exactly zero, not reduced, not delayed, right down to zero intensity worth of effect.

11.4 Stopping Potential

The stopping potential $V_0$ is the retarding (reversed) voltage applied between the emitting surface and a collector that is just large enough to stop even the fastest photoelectrons from reaching the collector, reducing the photocurrent to exactly zero. Since these fastest electrons carry $KE_{\max}$ :

Stopping potential

eV_0 = KE_{\max}

Measuring $V_0$ at several frequencies and plotting it against $f$ gives a straight line of slope $h/e$ , one of the historic ways Planck’s constant itself was measured. Because $KE_{\max}$ depends only on frequency, $V_0$ does too, it does not change when intensity is changed at fixed frequency.

Worked example

Stopping potential from frequency and work function

A metal of work function 2.0 eV is illuminated with light of frequency 7×10¹⁴ Hz. Find the stopping potential.

hf = (4.1357\times10^{-15})(7\times10^{14}) \approx 2.895\text{ eV}

KE_{\max} = 2.895 - 2.0 \approx 0.895\text{ eV} \;\Rightarrow\; V_0 = \textcolor{#e08a1e}{0.90\text{ V}}

(Since $eV_0=KE_{\max}$ , the stopping potential in volts is numerically equal to $KE_{\max}$ in eV.)

11.5 De Broglie Wavelength of Matter

De Broglie extended wave-particle duality the other way: if radiation (a wave) can behave like particles (photons), perhaps matter (particles) can behave like waves. He proposed that any particle of momentum $p=mv$ has an associated wavelength:

De Broglie wavelength

\lambda = \dfrac{h}{p} = \dfrac{h}{mv}

For an electron of charge $e$ accelerated from rest through a potential difference $V$ , its kinetic energy is $KE=eV$ , so its momentum is $p=\sqrt{2m\,KE}=\sqrt{2meV}$ , giving:

De Broglie wavelength after acceleration through V

\lambda = \dfrac{h}{\sqrt{2meV}}

Worked example

De Broglie wavelength of an accelerated electron

Find the de Broglie wavelength of an electron accelerated from rest through a potential difference of 100 V. (h = 6.626×10⁻³⁴ J·s, m = 9.11×10⁻³¹ kg, e = 1.602×10⁻¹⁹ C)

p = \sqrt{2meV} = \sqrt{2(9.11\times10^{-31})(1.602\times10^{-19})(100)} \approx 5.40\times10^{-24}\text{ kg·m/s}

\lambda = \dfrac{h}{p} = \dfrac{6.626\times10^{-34}}{5.40\times10^{-24}} \approx \textcolor{#e08a1e}{1.23\times10^{-10}\text{ m}\ (1.23\text{ Å})}

This is comparable to the spacing between atoms in a crystal, which is exactly why electrons accelerated through voltages like this can diffract off a crystal lattice, just like X-rays.

11.6 Davisson-Germer Experiment

In 1927, Davisson and Germer fired a beam of electrons at a nickel crystal and observed a diffraction pattern, intensity peaks at specific angles, matching almost exactly the de Broglie wavelength predicted for those electrons’ momentum. This was the first direct experimental confirmation that matter itself has wave nature, not just a theoretical proposal.

— From the NCERT exercises

Two of the chapter’s own exercise questions, with original worked solutions, both exercising the photon-momentum result of §11.2.1.

NCERT Exercise 11.4

Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam. (b) How many photons per second, on average, arrive at a target irradiated by this beam? (c) How fast does a hydrogen atom have to travel to have the same momentum as that of the photon?

Solution

(a) Energy and momentum of each photon, from §11.2.1:

f=\dfrac{c}{\lambda}=\dfrac{3\times10^8}{632.8\times10^{-9}}\approx4.74\times10^{14}\text{ Hz}, \qquad E=hf\approx\textcolor{#e08a1e}{3.14\times10^{-19}\text{ J}}

p=\dfrac{h}{\lambda}=\dfrac{6.63\times10^{-34}}{632.8\times10^{-9}}\approx\textcolor{#e08a1e}{1.05\times10^{-27}\text{ kg·m/s}}

(b) Power is energy per photon times photons per second:

N=\dfrac{P}{E}=\dfrac{9.42\times10^{-3}}{3.14\times10^{-19}}\approx\textcolor{#e08a1e}{3.0\times10^{16}\text{ photons/s}}

(c) Setting a hydrogen atom’s momentum (mass $\approx1.66\times10^{-27}\text{ kg}$ ) equal to the photon’s:

v=\dfrac{p}{m}=\dfrac{1.05\times10^{-27}}{1.66\times10^{-27}}\approx\textcolor{#e08a1e}{0.63\text{ m/s}}

NCERT Exercise 11.11

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Solution

A photon’s own de Broglie wavelength, from §11.5, is $\lambda_{dB}=h/p$ . Substituting the photon momentum just derived in §11.2.1, $p=hf/c$ :

\lambda_{dB}=\dfrac{h}{p}=\dfrac{h}{hf/c}=\dfrac{c}{f}

and $c/f$ is, by definition, exactly the wavelength $\lambda$ of the electromagnetic wave itself. So $\textcolor{#e08a1e}{\lambda_{dB}=\lambda}$ : for light, de Broglie’s formula doesn’t predict a new, independent wavelength, it simply reproduces the wavelength the photon already had, exactly as it must for the hypothesis to be a genuine generalisation rather than a contradiction.