Dual Nature of Radiation & Matter

The photoelectric effect, first observed by Hertz (1887) and studied experimentally by Lenard, is the emission of electrons from a metal surface when light of sufficiently high frequency falls on it. Key experimental observations: (1) Emission is instantaneous (no time lag, even at low intensity). (2) Maximum kinetic energy of photoelectrons depends on FREQUENCY, not intensity. (3) Intensity affects only the number of photoelectrons (photocurrent magnitude). (4) Below a threshold frequency $\nu_{0}$, no emission occurs regardless of intensity.

Einstein's photoelectric equation explains all observations using the photon concept: $\text{KE}_{max}$ = h $\nu$ - $\phi$ = h($\nu$ - $\nu_{0}$), where h = 6.63 x $10^{-34}$ J s is Planck's constant; [h] = [M $L^{2} \; T^{-1}$], SI unit: J s; $\phi$ = h $\nu_{0}$ is the work function (minimum energy to release an electron from the surface); [$\phi$] = [M $L^{2} \; T^{-2}$], SI unit: J or eV; and $\nu_{0}$ is the threshold frequency.

The stopping potential $V_{0}$ is the retarding potential that stops the most energetic photoelectrons: eV_0 = $\text{KE}_{max}$ = h $\nu$ - $\phi$, giving $V_{0}$ = (h/e) $\nu$ - $\phi$/e; [$V_{0}$] = [M $L^{2} \; T^{-3} \; A^{-1}$], SI unit: volt (V). Crucially, $V_{0}$ depends on frequency, NOT intensity. Higher intensity means more photons and higher saturation current, but the same $V_{0}$.

Three key graphs: (i) $\text{KE}_{max}$ vs $\nu$ — straight line, slope = h (universal, same for all metals), x-intercept = $\nu_{0}$, y-intercept = -$\phi$. (ii) Photocurrent I vs V — saturation at large positive V, reaches zero at -$V_{0}$; different intensities give different saturation currents but same $V_{0}$.
(iii) $V_{0}$ vs $\nu$ — straight line, slope = h/e, x-intercept = $\nu_{0}$.

A photon has energy E = h $\nu$ = hc/$\lambda$; [E] = [M $L^{2} \; T^{-2}$], SI unit: J (or eV, where 1 eV = 1.6 x $10^{-19}$ J).
Photon momentum: p = h/$\lambda$ = E/c = h $\nu$/c; [p] = [M L $T^{-1}$], SI unit: kg m/s. A photon has zero rest mass and always travels at speed c.

de Broglie hypothesis: Every moving particle has an associated wavelength $\lambda$ = h/(mv) = h/p, where m is mass and v is velocity; [$\lambda$] = [L], SI unit: m.
For an electron accelerated through potential V: KE = eV = ($\frac{1}{2}$)$mv^{2}$, so p = $\sqrt{2meV}$ and $\lambda$ = h/$\sqrt{2meV}$.
Numerically: $\lambda$ = 1.227/$\sqrt{V}$ nm (valid for electrons only).
Other forms: $\lambda$ = h/$\sqrt{2mKE}$ (from kinetic energy) and $\lambda$ = h/$\sqrt{3mkT}$ (thermal de Broglie wavelength, where k = 1.38 x $10^{-23}$ J/K is Boltzmann's constant).

The Davisson-Germer experiment (1927) provided direct confirmation of de Broglie's hypothesis by observing electron diffraction from a nickel crystal. The diffraction pattern matched the predicted de Broglie wavelength, proving that electrons exhibit wave behavior.

The key testable concept is Einstein's photoelectric equation ($\text{KE}_{max}$ = h $\nu$ - $\phi$) and its graphical interpretations, combined with de Broglie wavelength calculations for accelerated electrons.

Solved Numericals

N₁. Light of wavelength 400 nm strikes a metal surface with work function 2.0 eV. Find: (a) energy of incident photon in eV, (b) maximum KE of photoelectrons, (c) stopping potential.

Given: $\lambda$ = 400 nm = 400 x $10^{-9}$ m = 4 x $10^{-7}$ m, $\phi$ = 2.0 eV, h = 6.63 x $10^{-34}$ J s, c = 3 x $10^{8}$ m/s, 1 eV = 1.6 x $10^{-19}$ J.

(a) Photon energy: E = hc/$\lambda$ = (6.63 x $10^{-34}$ J s x 3 x $10^{8}$ m/s) / (4 x $10^{-7}$ m) = 19.89 x $10^{-26}$ / 4 x $10^{-7}$ J = 4.97 x $10^{-19}$ J = 4.97 x $10^{-19}$ / 1.6 x $10^{-19}$ eV = 3.11 eV.

(b) Maximum KE: $\text{KE}_{max}$ = E - $\phi$ = 3.11 eV - 2.0 eV = 1.11 eV.

In joules: $\text{KE}_{max}$ = 1.11 x 1.6 x $10^{-19}$ = 1.776 x $10^{-19}$ J.

(c) Stopping potential: eV_0 = $\text{KE}_{max}$ $V_{0}$ = $\text{KE}_{max}$/e = 1.11 eV / e = 1.11 V.

Note: Since the photon energy (3.11 eV) exceeds the work function (2.0 eV), the frequency is above threshold and emission occurs.

N₂. Find the de Broglie wavelength of an electron accelerated through 100 V. Compare with a proton through the same potential.

For electron: $m_{e}$ = 9.1 x $10^{-31}$ kg, e = 1.6 x $10^{-19}$ C, V = 100 V.

$\lambda_{e}$ = h/$\sqrt{2m_e eV}$ = 6.63 x $10^{-34}$ / $\sqrt{2 x 9.1 x 10^{-31} x 1.6 x 10^{-19} x 100}$ = 6.63 x $10^{-34}$ / $\sqrt{2 x 9.1 x 1.6 x 10^{-48}}$ = 6.63 x $10^{-34}$ / $\sqrt{29.12 x 10^{-48}}$ = 6.63 x $10^{-34}$ / (5.396 x $10^{-24}$) = 1.229 x $10^{-10}$ m = 0.123 nm = 1.23 angstrom.

Using shortcut: $\lambda$ = 1.227/$\sqrt{V}$ nm = 1.227/$\sqrt{100}$ = 1.$\frac{227}{10}$ = 0.1227 nm. Matches.

For proton: $m_{p}$ = 1.67 x $10^{-27}$ kg.

$\lambda_{p}$ = h/$\sqrt{2m_p eV}$ = 6.63 x $10^{-34}$ / $\sqrt{2 x 1.67 x 10^{-27} x 1.6 x 10^{-19} x 100}$ = 6.63 x $10^{-34}$ / $\sqrt{5.344 x 10^{-44}}$ = 6.63 x $10^{-34}$ / (2.312 x $10^{-22}$) = 2.868 x $10^{-12}$ m = 0.00287 nm.

Ratio: $\frac{\lambda_{e}}{\lambda_{p}}$ = $\sqrt{{m}_{p}/{m}_{e}}$ = $\sqrt{1.67 x 10^{-27} / 9.1 x 10^{-31}}$ = $\sqrt{1835}$ = 42.8.

The electron has ~43 times longer wavelength than the proton at the same accelerating potential, because $\lambda$ proportional to 1/$\sqrt{m}$.

N₃. The threshold wavelength for a metal is 5000 angstrom. Find the work function in eV and the stopping potential when light of wavelength 4000 angstrom is used.

Given: $\lambda_{0}$ = 5000 A = 5000 x $10^{-10}$ m = 5 x $10^{-7}$ m.

Work function: $\phi$ = hc/$\lambda_{0}$ = (6.63 x $10^{-34}$ x 3 x $10^{8}$) / (5 x $10^{-7}$) = 19.89 x $10^{-26}$ / 5 x $10^{-7}$ J = 3.978 x $10^{-19}$ J = 3.978 x $10^{-19}$ / 1.6 x $10^{-19}$ eV = 2.49 eV.

Incident light: $\lambda$ = 4000 A = 4 x $10^{-7}$ m. Photon energy: E = hc/$\lambda$ = 19.89 x $10^{-26}$ / 4 x $10^{-7}$ = 4.97 x $10^{-19}$ J = 3.11 eV.

$\text{KE}_{max}$ = E - $\phi$ = 3.11 - 2.49 = 0.62 eV.

Stopping potential: $V_{0}$ = $\text{KE}_{max}$/e = 0.62 V.

Alternatively using wavelengths: $\text{KE}_{max}$ = hc(1/$\lambda$ - 1/$\lambda_{0}$) = hc($\frac{1}{4000}$ - $\frac{1}{5000}$) $A^{-1}$ = hc x (5000 - 4000)/(4000 x 5000) $A^{-1}$ = hc/(20000 A) = hc/(2 x $10^{-6}$ m). This gives the same result: 9.945 x $10^{-20}$ J = 0.62 eV.