Answer:
(c) 6.91x10^14 Hz
Explanation:
Find the level energy of n=2 and n=5, using the formula:
[tex]E = -E_0/n^2[/tex]
where [tex]E_0=13.6eV[/tex]
[tex]E_2 =\frac{-13.6}{2^2}=-3.4eV[/tex]
[tex]E_5 =\frac{-13.6}{5^2}=-0.544eV[/tex]
To jump from n=2 to n=5 the electron absorbs a photon with energy equal to [tex](-0.544) - (-3.4)=2.856eV[/tex], using the next formula to find specific wavelength [tex]\lambda[/tex] to that energy
[tex]E = hc/\lambda[/tex]
Where [tex]c[/tex] is the speed of light ([tex]c=3 \times10^8m/s[/tex]) and [tex]h[/tex] is Planck's constant ([tex]h=4.14\times10^{-15}eVs[/tex]). Solve for [tex]\lambda[/tex]:
[tex]E = hc/\lambda\\\lambda E = hc\\\lambda = \frac{hc}{E} \\\lambda = \frac{(4.14\times10^{-15})(3 \times10^8)}{2.856}=4.35\times10^{-7}m[/tex]
The frequency of this wavelength is calculated with this formula:
[tex]f=\frac{c}{\lambda}[/tex]
[tex]f=\frac{3\times10^8}{4.3487\times10^{-7}} =6.89\times10^{14}Hz\approx6.9\times10^{14}Hz[/tex]
