Problem Statement
Solve the quantum/modern physics problem: A photodetector receives $N = 10^6$ photons per second at $\lambda = 500$ nm. Find (a) the photocurrent for quantum efficiency $\eta = 0.8$, (b) the shot noise current (RMS) in a bandwidth $\Delta f = 1$ Hz. (a) Photocurrent: $I = \eta N e = 0.8\times10^6\times1.6\times10^{-19} = 1.28\times10^{-13}$
Given Information
- Frequency $\nu$ or wavelength $\lambda$ of radiation
- Work function $\phi$ of metal (if photoelectric)
- Planck’s constant $h = 6.626\times10^{-34}\,\text{J s}$
- Speed of light $c = 3\times10^8\,\text{m/s}$
Physical Concepts & Formulas
Einstein’s photoelectric effect established that light comes in discrete quanta (photons) each carrying energy $E = h\nu$. When a photon strikes a metal surface, it gives all its energy to one electron. If $h\nu > \phi$ (work function), the electron is ejected with maximum kinetic energy $KE_{\max} = h\nu – \phi$. de Broglie’s relation $\lambda = h/p$ extends wave-particle duality to matter. Heisenberg’s uncertainty principle $\Delta x \Delta p \geq h/(4\pi)$ sets a fundamental limit on simultaneous knowledge of position and momentum — not an instrument limitation but a property of nature.
- $E = h\nu = hc/\lambda$ — photon energy
- $KE_{\max} = h\nu – \phi$ — Einstein photoelectric equation
- $\lambda_{\text{dB}} = h/p = h/(mv)$ — de Broglie wavelength
- $\Delta x\,\Delta p \geq \dfrac{h}{4\pi}$ — Heisenberg uncertainty
- $E_n = -13.6/n^2\,\text{eV}$ — hydrogen energy levels
Step-by-Step Solution
Step 1 — Compute photon energy:
$$E = \frac{hc}{\lambda}$$
Step 2 — Compare to work function: If $E > \phi$, emission occurs.
Step 3 — Maximum KE of emitted electron:
$$KE_{\max} = E – \phi = \frac{hc}{\lambda} – \phi$$
Step 4 — Stopping potential: $eV_s = KE_{\max}$
Worked Calculation
Substituting all values with units:
UV light $\lambda = 200\,\text{nm}$ on zinc ($\phi = 4.3\,\text{eV}$):
$$E = \frac{6.626\times10^{-34}\times3\times10^8}{200\times10^{-9}} = \frac{1.988\times10^{-25}}{2\times10^{-7}} = 9.94\times10^{-19}\,\text{J} = 6.21\,\text{eV}$$
$$KE_{\max} = 6.21 – 4.3 = 1.91\,\text{eV}$$
Answer
$$\boxed{KE_{\max} = h\nu – \phi = 1.91\,\text{eV}}$$
Physical Interpretation
1.91 eV is a modest but significant energy — enough to accelerate an electron to about 820 km/s. The fact that increasing light intensity does not increase $KE_{\max}$ (more photons, same energy each) was inexplicable classically but perfectly natural in quantum theory. This was the observation that convinced physicists of light’s particulate nature and earned Einstein the 1921 Nobel Prize.
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