Problem Statement
Solve the quantum/modern physics problem: Solve the work-energy problem: Find the mean energy of photons in blackbody radiation at $T=300\ \text{K}$. The mean photon energy in blackbody radiation: $$\langle E\rangle = \frac{\int_0^\infty h\nu\,u(\nu)\,d\nu}{\int_0^\infty u(\nu)\,d\nu} = \frac{\sigma_{SB}T^4/c \cdot 4}{\text{(number density)
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\langle E\rangle = \frac{\int_0^\infty h\nu\,u(\nu)\,d\nu}{\int_0^\infty u(\nu)\,d\nu} = \frac{\sigma_{SB}T^4/c \cdot 4}{\text{(number density)
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:
$$
$$
Step 2 — Compare to work function: If $E > \phi$, emission occurs.
Step 3 — Maximum KE of emitted electron:
$$
$$
Step 4 — Stopping potential: $eV_s = KE_{\max}$
Worked Calculation
Substituting all values with units:
UV light $\lambda = 200\,\text{nm}$ on zinc ($\phi = 4.3\,\text{eV}$):
$$
Answer
$$\boxed{KE_{\max} = h\nu – \phi = 1.91\,\text{eV}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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