Answer
The multiplication factor is $~~4560$
Work Step by Step
We can find the change in energy when the electron jumps from $n=3$ to $n=2$:
$E = (-\frac{13.6~eV}{2^2})-(-\frac{13.6~eV}{3^2})$
$E = -1.89~eV$
To conserve energy, a photon with an energy of $E = 1.89~eV$ is released.
We can find the original wavelength of this photon:
$E = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.89~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda = 6.57\times 10^{-7}~m$
We can find the multiplication factor:
$\frac{3.00\times 10^{-3}~m}{6.57\times 10^{-7}~m} = 4560$
The multiplication factor is $~~4560$
