Exercises — EM spectrum — all bands and applications

Energy grows with frequency, and frequency grows as we walk this list. So, low → high energy: $\text{radio}<\text{microwave}<\text{infrared}<\text{visible}<\text{ultraviolet}<\text{X-ray}<\text{gamma}.$ Why this order? Each step to the right shortens the wavelength; a shorter wave crams more crests into each second, so $\nu$ rises, and $E=h\nu$ rises with it. Nothing to memorise beyond the one chain $\lambda \downarrow \Rightarrow \nu \uparrow \Rightarrow E\uparrow$.

• $3\text{cm}=0.03\text{m}$ sits in the range $1\text{mm}$–$0.1\text{m}$ → microwave (radar band).
• $500\text{nm}$ is between $400$ and $700\text{nm}$ → visible (blue-green).
• $0.05\text{nm}<0.01\text{nm}$? No — $0.05>0.01$, so it is in $0.01$–$10\text{nm}$ → X-ray (near the atomic-spacing scale that makes crystals diffract).

Which tool and why? We are given frequency and want length, and the two are locked by Wave speed c = νλ. Rearrange for $\lambda$: $\lambda =\frac{c}{\nu }=\frac{3\times 10^8\text{m/s}}{5\times 10^9\text{Hz}}=6\times 10^{-2}\text{m}=6\text{cm}.$ Sanity check: $6\text{cm}$ lands in the microwave band — exactly where Wi-Fi lives.

Which tool and why? We want per-photon energy from a wavelength, so use Photon energy E = hν in its wavelength form $E=hc/\lambda$ (because $\nu =c/\lambda$). $E=\frac{hc}{\lambda }=\frac{(6.63\times 10^{-34}\text{J⋅s})(3\times 10^8\text{m/s})}{650\times 10^{-9}\text{m}}=3.06\times 10^{-19}\text{J}.$ Convert to eV by dividing by the "one-eV in joules" exchange rate: $E=\frac{3.06\times 10^{-19}\text{J}}{1.6\times 10^{-19}\text{J/eV}}\approx 1.91\text{eV}.$ Sanity check: a couple of eV is the scale of Atomic spectra — which is why atoms emit visible light.

Total energy $=$ (photons) $\times$ (energy per photon). First the per-photon energy: $E_1=\frac{hc}{\lambda }=\frac{(6.63\times 10^{-34}\text{J⋅s})(3\times 10^8\text{m/s})}{550\times 10^{-9}\text{m}}=3.616\times 10^{-19}\text{J}.$ Then divide the pulse energy by one photon's worth: $N=\frac{2.0\times 10^{-3}\text{J}}{3.616\times 10^{-19}\text{J}}\approx 5.5\times 10^{15}\text{photons}.$ Why divide? Each photon is one indivisible lump; the count is simply how many lumps fit into the total.

Since $E=hc/\lambda$, the constants $hc$ cancel in a ratio — that is why ratios are the fast route: $\frac{E_A}{E_B}=\frac{hc/{\lambda }_A}{hc/{\lambda }_B}=\frac{{\lambda }_B}{{\lambda }_A}.$ Put both wavelengths in metres: ${\lambda }_A=0.1\times 10^{-9}\text{m}=1\times 10^{-10}\text{m}$, and ${\lambda }_B=10\times 10^{-6}\text{m}=1\times 10^{-5}\text{m}$. $\frac{E_A}{E_B}=\frac{1\times 10^{-5}\text{m}}{1\times 10^{-10}\text{m}}=1\times 10^5.$ Interpretation: the X-ray photon is $100,000\times$ more energetic — enough to ionise atoms and pass through soft tissue, while the IR photon merely sets molecules vibrating (warmth). Same kind of wave, wildly different consequence.

"Longest wavelength that still works" = "exactly the threshold energy", because longer $\lambda$ means less energy and would fail. So set $E=5\text{eV}$ and solve $E=hc/\lambda$ for $\lambda$. First convert to joules: $E=5\text{eV}\times 1.6\times 10^{-19}\text{J/eV}=8.0\times 10^{-19}\text{J}$. $\lambda =\frac{hc}{E}=\frac{(6.63\times 10^{-34}\text{J⋅s})(3\times 10^8\text{m/s})}{8.0\times 10^{-19}\text{J}}\approx 2.49\times 10^{-7}\text{m}=249\text{nm}.$ Which band? $249\text{nm}$ is below $400\text{nm}$ → ultraviolet. That is precisely why UV sterilises (breaks DNA bonds) and visible light does not.

The figure below plots $E=hc/\lambda$ on log axes. Follow the blue curve down-and-right: as wavelength grows, energy falls. The yellow dashed line marks the $5\text{eV}$ bond threshold; it meets the curve (green dot) exactly at the red dashed wavelength, $\approx 249\text{nm}$ — the border between the shaded UV and visible bands. Anything to the right of that red line is too weak to break the bond.

Which tool and why? Bragg's law relates the path-length difference between waves bouncing off successive planes to the wavelength; constructive interference needs that extra path to be a whole number of wavelengths. Solve for $\sin \theta$: $\sin \theta =\frac{n\lambda }{2d}=\frac{(1)(0.154\text{nm})}{2(0.282\text{nm})}=\frac{0.154}{0.564}=\mathrm{0.2730.}$ Then $\theta =\arcsin (0.2730)\approx 15.85^{\circ }$. Why arcsin? We know the sine (a ratio) and want the angle that produces it — arcsin is exactly the "which angle has this sine?" undo-operation.

(a) $\lambda =c/\nu =\frac{3\times 10^8\text{m/s}}{90\times 10^6\text{Hz}}=3.33\text{m}$. A quarter of that is $\frac{3.33\text{m}}{4}=0.833\text{m}\approx 83\text{cm}$. (This is why real FM whip antennas are order-of-a-metre long.) (b) ${\lambda }_{\text{oven}}=\frac{3\times 10^8\text{m/s}}{2.45\times 10^9\text{Hz}}=0.1224\text{m}\approx 12.2\text{cm}$. (c) Energy ratio $=\frac{{\lambda }_{\text{oven}}}{{\lambda }_{\text{vis}}}=\frac{0.1224\text{m}}{500\times 10^{-9}\text{m}}=\frac{0.1224}{5\times 10^{-7}}\approx 2.45\times 10^5$. So a single green photon is about $245,000\times$ more energetic than a microwave-oven photon — yet the oven cooks food and the green light does not, because cooking is about total power delivered and dielectric relaxation, not per-photon punch.

Oven photon: $E=h\nu =(6.63\times 10^{-34}\text{J⋅s})(2.45\times 10^9\text{Hz})=1.62\times 10^{-24}\text{J}$. In eV: $E=\frac{1.62\times 10^{-24}\text{J}}{1.6\times 10^{-19}\text{J/eV}}=1.01\times 10^{-5}\text{eV}$. Compare to the transition: $\frac{2\times 10^{-3}\text{eV}}{1.01\times 10^{-5}\text{eV}}\approx 198$. The transition needs roughly $200\times$ more energy than one oven photon supplies. Conclusion: the oven can't lift a molecule across that gap with one photon — confirming the mechanism is bulk dielectric relaxation (the field drags polar molecules and their lag dumps heat), not a discrete resonance. This is exactly the steel-manned mistake from the parent note, made quantitative.

(a) Refractive index is defined by $n=c/v$, so $v=c/n=\frac{3\times 10^8\text{m/s}}{1.53}=1.96\times 10^8\text{m/s}$. Slower than vacuum — as it must be for $n>1$. (b) Frequency is set by the source and does not change on entering a medium (the wave crests arrive at the boundary at the same rate they leave it). With $v=\nu {\lambda }_{\text{glass}}$ and $\nu =c/{\lambda }_0$: ${\lambda }_{\text{glass}}=\frac{v}{\nu }=\frac{{\lambda }_0}{n}=\frac{400\text{nm}}{1.53}\approx 261\text{nm}.$ (c) No — frequency is unchanged (that is why colour is preserved). (d) No — $E=h\nu$, and $\nu$ is unchanged, so photon energy is identical in glass and vacuum. Only the speed and wavelength shrink. This is the resolution of the parent-note mistake "different bands travel at different speeds": speed changes happen in media, and even then energy stays fixed.

Which tool and why? We are given a temperature and asked where the emission is strongest — that is precisely the question Wien's law answers. It says the peak wavelength is inversely proportional to temperature, ${\lambda }_{\text{peak}}=b/T$: hotter body ⇒ shorter peak wavelength. So we simply feed each $T$ into that formula. (a) ${\lambda }_{\text{peak}}=\frac{2.90\times 10^{-3}\text{m⋅K}}{5800\text{K}}=5.0\times 10^{-7}\text{m}=500\text{nm}$ → visible (green-ish). (b) ${\lambda }_{\text{peak}}=\frac{2.90\times 10^{-3}\text{m⋅K}}{310\text{K}}=9.35\times 10^{-6}\text{m}\approx 9.4\mu \text{m}$ → infrared. (c) The Sun peaks in the visible, so evolution tuned our eyes to the band where the most photons arrive — that is why "visible" is nothing special physically, just the Sun's peak. Warm bodies near room temperature peak in the IR, so a camera that "sees" IR detects the glow every warm object emits, even in the dark. One law, two temperatures, two bands — the whole spectrum logic in miniature.

The figure below draws both blackbody curves on a shared log-wavelength axis. The yellow curve (Sun) peaks at its dashed line inside the shaded green visible band; the red curve (human) peaks far to the right at $\approx 9.4\mu \text{m}$, deep in the IR. Read off directly: cooler body ⇒ its whole curve — and its peak — slides rightward to longer wavelengths.

Antenna at 90 MHz quarter-wave length? ::: about 0.83 m Wi-Fi 5 GHz wavelength? ::: 6 cm 650 nm photon energy in eV? ::: about 1.91 eV Longest wavelength that breaks a 5 eV bond? ::: about 249 nm (UV) Bragg angle for 0.154 nm X-rays off 0.282 nm planes? ::: about 15.85° Does frequency change entering glass? ::: No — only speed and wavelength shrink Sun's blackbody peak wavelength and band? ::: 500 nm, visible