Exercises — Black-body radiation and Planck's quantum hypothesis E = hν

What we do: apply $E=h\nu$. Why this tool: the question hands us a frequency and asks for the smallest energy bundle — that is exactly what $E=h\nu$ delivers. $E=(6.626\times 10^{-34})(1.2\times 10^{15})=7.95\times 10^{-19}\text{J}$ Sanity check: higher $\nu$ than green light (Example 1 of the parent gave $3.3\times 10^{-19}$ J at $5\times 10^{14}$ Hz), so a bigger energy is expected. ✓

What we do: use $E=hc/\lambda$ (combines $\nu =c/\lambda$ and $E=h\nu$ in one step). Why this tool: we're given wavelength, not frequency, so the wavelength form saves a conversion. Convert nm to m: $650\text{nm}=650\times 10^{-9}\text{m}$. $E=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{650\times 10^{-9}}=3.06\times 10^{-19}\text{J}$ Sanity check: red light (long $\lambda$, low $\nu$) should carry less energy than UV. It does. ✓

Step 1 — one quantum: $E_1=hc/\lambda$. $E_1=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{500\times 10^{-9}}=3.976\times 10^{-19}\text{J}$ Step 2 — scale to a mole. Why? A mole is just $N_A$ copies, so multiply. $E=E_1{N}_A=(3.976\times 10^{-19})(6.022\times 10^{23})=2.394\times 10^5\text{J/mol}$ $=239\text{kJ/mol}$ Why this matters: ~240 kJ/mol sits right in the range of chemical bond strengths — visible light can drive photochemistry.

Step 1 — energy of one photon: $E_1=hc/\lambda$. $E_1=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{550\times 10^{-9}}=3.614\times 10^{-19}\text{J}$ Step 2 — divide power by energy-per-photon. Why? Power is joules per second; each photon carries $E_1$ joules; so the count per second is total energy per second ÷ energy each. $\frac{2.0\times 10^{-3}\text{J/s}}{3.614\times 10^{-19}\text{J}}=5.53\times 10^{15}\text{photons/s}$ Sanity check: a tiny power still spits out quadrillions of photons — because each packet is minuscule. This is exactly why energy looks continuous.

Why $\frac{h\nu }{kT}$? It is the only dimensionless knob in Planck's average energy $⟨E⟩=\frac{h\nu }{e^{h\nu /kT}-1}$. Small ratio → classical ($⟨E⟩\to kT$); large ratio → frozen ($⟨E⟩\to 0$). Look at the red curve in the figure.

First $kT=(1.381\times 10^{-23})(300)=4.143\times 10^{-21}\text{J}$.

IR mode: $\frac{h{\nu }_1}{kT}=\frac{(6.626\times 10^{-34})(3\times 10^{12})}{4.143\times 10^{-21}}=0.480$. Since $0.48<1$, one step is smaller than the thermal budget — the mode is classically alive. Check: $⟨E⟩=\frac{h{\nu }_1}{e^{0.48}-1}=\frac{1.988\times 10^{-21}}{0.616}=3.23\times 10^{-21}\text{J}\approx 0.78kT$. Close to $kT$. ✓

UV mode: $\frac{h{\nu }_2}{kT}=\frac{(6.626\times 10^{-34})(3\times 10^{15})}{4.143\times 10^{-21}}=480$. Since $480\gg 1$, one step is hundreds of times the thermal budget — the mode is frozen out. Check: $e^{480}$ is astronomically large, so $⟨E⟩\approx h{\nu }_2{e}^{-480}\approx 0\text{J}$. ✓

The lesson: at 300 K the UV modes get essentially nothing — that is precisely how Planck kills the ultraviolet catastrophe.

Classical: $⟨E{⟩}_{\text{cl}}=kT=4.143\times 10^{-21}\text{J}$ (equipartition gives every mode $kT$). Planck: $⟨E⟩=\frac{h{\nu }_2}{e^{480}-1}\approx h{\nu }_2{e}^{-480}$. Numerically $e^{-480}\approx 10^{-208}$, so $⟨E⟩\approx (1.988\times 10^{-18})(10^{-208})\approx 10^{-226}\text{J}$ — effectively zero. Overestimate factor: $\frac{kT}{⟨E⟩}\approx \frac{4\times 10^{-21}}{10^{-226}}\approx 10^{205}$. Interpretation: classical physics assigns $kT$ to a mode that in truth holds nothing. Summed over infinitely many such UV modes, that error diverges — the catastrophe. Planck's exponential cutoff is what saves reality.

(a) Wien's Displacement Law: ${\lambda }_{max}T=b\Rightarrow T=\frac{b}{{\lambda }_{max}}$. Why this law? It links the observed peak colour directly to temperature. $T=\frac{2.898\times 10^{-3}}{480\times 10^{-9}}=6038\text{K}$ (b) Peak-photon energy $E=hc/{\lambda }_{max}$. $E=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{480\times 10^{-9}}=4.14\times 10^{-19}\text{J}$ (c) Thermal energy $kT=(1.381\times 10^{-23})(6038)=8.34\times 10^{-20}\text{J}$. Ratio $\frac{E}{kT}=\frac{4.14\times 10^{-19}}{8.34\times 10^{-20}}=4.96\approx 5$. Insight: the peak of a black-body spectrum always sits near $h\nu \approx 5kT$ — the sweet spot where steps are neither too tall to climb nor so short that few modes exist. Every black body, from a furnace to a star, obeys this same ratio.

Why Stefan–Boltzmann? It gives the total area under the intensity curve, i.e. all the energy at every wavelength summed. Power scales as $T^4$, so $\frac{P_2}{P_1}={\left(\frac{T_2}{T_1}\right)}^4={\left(\frac{2000}{1000}\right)}^4=2^4=16$ Insight: doubling temperature multiplies the glow 16-fold — a small heating gives a huge brightness jump. This is why a filament leaps from dull red to dazzling white over a modest temperature range.

Step 1 — name the small quantity. Let $x=\frac{h\nu }{kT}$. "$h\nu \ll kT$" means $x\ll 1$. Step 2 — Taylor-expand the exponential. Why? For small $x$, $e^x\approx 1+x+\frac{x^2}{2}$. This is legitimate only because $x$ is small (echoing the L3 warning). $e^x-1\approx x+\frac{x^2}{2}\approx x(\text{keep leading term})$ Step 3 — substitute back. $⟨E⟩=\frac{h\nu }{e^x-1}\approx \frac{h\nu }{x}=\frac{h\nu }{h\nu /kT}=kT$ So Planck's law contains the classical Rayleigh–Jeans result as its low-frequency shadow. ✓ Numerical check at $x=0.01$: $\frac{⟨E⟩}{kT}=\frac{x}{e^x-1}=\frac{0.01}{e^{0.01}-1}=\frac{0.01}{0.010050}=0.99500$ Within $0.5\%$ of 1 — the classical value is essentially exact when steps are tiny. ✓

(a) Photon. Invert $E=hc/\lambda$: ${\lambda }_{\text{photon}}=\frac{hc}{E}=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{3.0\times 10^{-19}}=6.63\times 10^{-7}\text{m}=663\text{nm}$ Visible red — consistent with its energy. (b) Electron speed from $E=\frac{1}{2}{m}_e{v}^2\Rightarrow v=\sqrt{2E/m_e}$. Why? de Broglie needs momentum $m_{e}v$, so we need $v$ first. $v=\sqrt{\frac{2(3.0\times 10^{-19})}{9.11\times 10^{-31}}}=\sqrt{6.586\times 10^{11}}=8.12\times 10^5\text{m/s}$ de Broglie wavelength: ${\lambda }_{\text{electron}}=\frac{h}{m_{e}v}=\frac{6.626\times 10^{-34}}{(9.11\times 10^{-31})(8.12\times 10^5)}=8.96\times 10^{-10}\text{m}=0.896\text{nm}$ Comment: at the same energy the photon's wave (663 nm) is far longer than the electron's (0.90 nm), because the electron carries that energy as motion of a massive particle — its momentum is large, and momentum is what shrinks a wavelength. This is the seed of wave–particle duality that grows into the Bohr Model of the Atom and beyond.

Which formula for wavelength-given energy of a photon? ::: $E=hc/\lambda$ Which dimensionless ratio decides if a mode is frozen? ::: $h\nu /kT$ Peak of a black-body spectrum sits near $h\nu \approx ?kT$ ::: about $5kT$ Doubling $T$ multiplies total radiated power by? ::: $2^4=16$ de Broglie wavelength of a massive particle? ::: $\lambda =h/(mv)$