The Light-Emitting Diode (LED) — Question 2

Question 2

A narrow energy spectrum maps to a wavelength width

An LED emits near $E_0=2.00\,\mathrm{eV}$ with small energy full width at half maximum $\Delta E=.0800\,\mathrm{eV}$. Use $hc=1240\,\mathrm{eV\,nm}$ and neglect line-shape changes to first order.

1. Derive the first-order relation between energy and wavelength widths.

2. Evaluate the central wavelength and wavelength width, then compare the exact wavelength interval corresponding to $E_0\pm\Delta E/2$.

Question 2 - Solution

A narrow energy spectrum maps to a wavelength width

(a) Method and calculation

$$\lambda=hc/E,\quad d\lambda/dE=-hc/E^2.$$$$\boxed{\Delta\lambda\simeq(hc/E_0^2)\Delta E,\quad\Delta\lambda/\lambda_0\simeq\Delta E/E_0}.$$

(b) Interpretation and further calculation

$$\boxed{\lambda_0=620\,\mathrm{nm},\quad\Delta\lambda\simeq24.8\,\mathrm{nm}}.$$$$\boxed{\lambda(1.96)-\lambda(2.04)=632.653-607.843=24.810\,\mathrm{nm}}.$$

Physical check

The exact interval is slightly asymmetric in wavelength. A spectral density also transforms with a Jacobian, so its exact wavelength FWHM need not equal this endpoint interval.