Exact and approximate relations between the bandwidth in wavelength and wavenumber units is given by: $$ \Delta\lambda = \frac{4\pi c}{\Delta \omega} \left( \sqrt{1+\frac{\lambda_0^2\Delta \omega^2}{4\pi^2 c^2}} - 1 \right) \approx \frac{\Delta \omega\lambda_0^2}{2\pi c} = \Delta k \lambda_0^2. $$
If bandwidth \( \Delta k \) is given in inverse centimeters, bandwidth in nanometers is approximately $$ \Delta\lambda\mathrm{[nm]} \approx 10^{-7} \cdot \Delta k\mathrm{[cm^{-1}]}\cdot(\lambda_0\mathrm{[nm]})^2. $$
Peak width relations: $$ \mathrm{FWHM} = 2\sqrt{2\ln2}\sigma,$$ $$D_{1/\mathrm{e}^2} = 4\sigma = \sqrt{\frac{2}{\ln2}}\mathrm{FWHM},$$ $$D_{1/\mathrm{e}} = 2\sqrt{2}\sigma = \frac{\mathrm{FWHM}}{\sqrt{\ln 2}}.$$