Answer
The corresponding frequency width for the emission is $~~7.49\times 10^9~Hz$
Work Step by Step
The light has a wavelength between $\lambda_1 = 632.795~nm$ and $\lambda_2 = 632.805~nm$
In general: $f~\lambda = c$
We can find the frequency associated with each wavelength:
$f_1 = \frac{3.0\times 10^8~m/s}{632.795\times 10^{-9}~m} = 4.74087185\times 10^{14}~Hz$
$f_2 = \frac{3.0\times 10^8~m/s}{632.805\times 10^{-9}~m} = 4.74079693\times 10^{14}~Hz$
We can find the width between these two frequencies:
$\Delta f = f_1 - f_2$
$\Delta f = 4.74087185\times 10^{14}~Hz-4.74079693\times 10^{14}~Hz$
$\Delta f = 7.49\times 10^9~Hz$
The corresponding frequency width for the emission is $~~7.49\times 10^9~Hz$.
