electromagnetic-wave emission and spectrum

Electromagnetic waves are emitted by accelerated charges: a charge at rest or moving uniformly emits no radiation, whereas charge oscillations in an antenna or electron transitions between levels produce changing fields that propagate as a wave. From the wave equation for the fields it follows that in vacuum all such waves travel at speed $c$, and frequency $\nu$, wavelength $\lambda$, and speed are related by $c=\lambda\nu.$ In a medium the speed decreases to $v=c/n$, so the wavelength changes, but the frequency remains that of the source — this is why the spectrum scale is conveniently organized by frequency, and the quantum energy is found as $E=h\nu=\frac{hc}{\lambda}.$ The electromagnetic spectrum includes radio, microwave, infrared, visible, ultraviolet, X-ray, and gamma radiation; the band boundaries are conventional, and their order is set by frequency. As with the waves studied earlier, propagation, interference, and diffraction confirm the wave nature of radiation: a diffraction grating separates light by wavelength, and interference enables highly accurate measurement of $\lambda$. Finally, thermal radiation from a hot body gives a continuous spectrum whose peak shifts to shorter wavelengths as temperature rises (Wien’s law), while rarefied gases give line spectra.

A radio station emits at frequency $\nu = 100$ MHz. Find the wavelength in vacuum.

Light of wavelength $\lambda = 500$ nm falls normally on a diffraction grating with spacing $d = 2{,}0$ µm. Find the angle at which the first-order maximum is observed ($m=1$).

A star has its thermal-radiation peak at wavelength $\lambda_{\max}=480$ nm. Estimate its surface temperature using Wien’s law $\lambda_{\max}T=b$, where $b=2{,}90\cdot10^{-3}$ m·K. Compare your result with the Sun’s temperature, for which $\lambda_{\max}\approx500$ nm.

A safe self-study investigation using a simulation. 1) Open the browser-based “Blackbody Spectrum” simulation (PhET) or a similar interactive thermal-radiation model. 2) Set several temperatures, for example $3000$ K, $4000$ K, $5000$ K, and $6000$ K, and record the peak wavelength $\lambda_{\max}$ for each. 3) Compute the product $\lambda_{\max}T$ and check that it is approximately Wien’s constant $b=2{,}90\cdot10^{-3}$ m·K. 4) Convert one of the found wavelengths to frequency using $\nu=c/\lambda$ and note which band of the electromagnetic spectrum it falls into. Do not use household microwave ovens, lasers, or UV sources for observations: perform all measurements only in the simulation. Write a conclusion about how the source temperature is related to the position of the peak on the spectrum scale.

Show the connection between the emission mechanism, frequency, wavelength, and the position of a band on the electromagnetic spectrum scale, and illustrate the shift of the thermal-radiation peak as temperature changes.

A diagram with three aligned parts: (1) an oscillating charge (dipole) with outgoing wavefronts and the label “accelerated charge → radiation”; (2) a horizontal logarithmic spectrum axis from radio to gamma radiation, labeled with frequencies $\nu$ and wavelengths $\lambda$, with the visible band highlighted and an arrow to the right corresponding to increasing frequency and photon energy $E=h\nu$; (3) a family of thermal-radiation curves for temperatures $3000$ K, $4000$ K, $5000$ K, and $6000$ K with marked maxima $\lambda_{\max}$ and a dashed Wien-law line $\lambda_{\max}T=b$.

A smartphone transmits at a frequency of 2.6 GHz. The speed of light is $c=3.00\times10^8$ m/s, so the wavelength is $\lambda=c/f=\frac{3.00\times10^8}{2.6\times10^9}\approx0.115$ m, about 11.5 cm. In 0.20 s the wave travels 60,000 km.

Accelerated electrons in the antenna create time-varying $E$ and $B$ fields; this is emission of an electromagnetic wave, and 2.6 GHz lies in the radio part of the spectrum. The relation $c=\lambda f$ connects frequency and wavelength.

The radar emits microwaves at 24.15 GHz. The wavelength is $\lambda=c/f=\frac{3.00\times10^8}{24.15\times10^9}\approx0.0124$ m $=1.24$ cm. Reflected from a car moving at 30 m/s, the signal changes frequency by $\Delta f \approx 2f v/c = \frac{2(24.15\times10^9)(30)}{3.00\times10^8}\approx 4.8$ kHz.

Emission by accelerated charges in the radar antenna produces microwaves; propagation at speed $c$ and the Doppler shift let the device measure speed.

In a microwave oven the magnetron produces waves at 2.45 GHz. The wavelength is $\lambda=c/f=\frac{3.00\times10^8}{2.45\times10^9}\approx0.122$ m $=12.2$ cm. The energy of one photon is $E=hf=(6.626\times10^{-34})(2.45\times10^9)\approx1.62\times10^{-24}$ J $\approx1.0\times10^{-5}$ eV. At 700 W for 60 s, 1000 g of water receives $Q=700\times60=4.2\times10^4$ J and warms by $\Delta T=Q/(m c_w)\approx10^{\circ}$C, where $c_w=4186$ J/(kg·K).

Microwaves are a region of the electromagnetic spectrum; absorption at 2.45 GHz is tied to rotational transitions of water molecules, and $E=hf$ shows that this photon does not ionize atoms.