Answer:
The higher energy level, the electron reached is 11
Explanation:
Given;
wavelength of the photon, λ = 91.94 nm
The energy of the absorbed photon is given by;
[tex]E = \frac{hc}{\lambda}\\\\ E = \frac{6.626*10^{-34}*3*10^8}{91.94*10^{-9}}\\\\E = 2.162*10^{-18} \ J[/tex]
At ground level, n₁ = 1
let the higher energy level = n₂
[tex]\delta E = -2.18*10^{-18}(\frac{1}{n_2^2}-\frac{1}{n_1^2} )\\\\ 2.162*10^{-18} = -2.18*10^{-18}(\frac{1}{n_2^2}-1 )\\\\\frac{ 2.162*10^{-18}}{-2.18*10^{-18}} = \frac{1}{n_2^2}-1\\\\-0.992 = \frac{1}{n_2^2}-1\\\\-0.992 + 1 = \frac{1}{n_2^2}\\\\0.008 = \frac{1}{n_2^2}\\\\n_2^2 = \frac{1}{0.008}\\\\ n_2 = \sqrt{\frac{1}{0.008}}\\\\ n_2 = \sqrt{125}\\\\n_2 = 11[/tex]
Therefore, the higher energy level, the electron reached is 11
