Problem 6.109 — Quantum Defect

Problem Statement

Solve the quantum/modern physics problem: Problem 6.109 — Quantum Defect See problem statement for all given quantities. This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention t

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.