Answer
The wavelength of the photon is $~~103~eV$
Work Step by Step
We can find the energy when $n=3$:
$E_3 = -\frac{13.60~eV}{3^2} = -1.511~eV$
We can find the energy when $n=1$:
$E_1 = -\frac{13.60~eV}{1^2} = -13.60~eV$
We can find the energy difference:
$E_3-E_1 = (-1.511~eV)-(-13.60~eV) = 12.1~eV$
The energy of the photon that is emitted is $~~12.1~eV$
We can find the wavelength associated with this energy:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{12.1~eV}$
$\lambda = 103\times 10^{-9}~m$
$\lambda = 103~nm$
The wavelength of the photon is $~~103~eV$
