Respuesta :
Explanation:
The given data is as follows.
[tex]n_{1}[/tex] = 9, [tex]n_{2}[/tex] = ?
Z for hydrogen = 1
As we know that,
Energy (E) = [tex]\frac{hc}{\lambda}[/tex]
where, h = planck's constant = [tex]6.626 \times 10^{-34}[/tex] Js
c = speed of light = [tex]3 \times 10^{8}[/tex] m/s
[tex]\lambda[/tex] = wavelength
According to Reydberg's equation, we will calculate the energy emitted by the photon as follows.
[tex]\Delta E = -2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
or, [tex]\frac{hc}{\lambda}[/tex] = [tex]-2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
Putting the given values into the above equation as follows.
[tex]\frac{hc}{\lambda}[/tex] = [tex]-2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}][/tex]
[tex]\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8}m/s}{384 \times 10^{-9} m}[/tex] = [tex]-2.179 \times 10^{-18} J \times (1)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(9)^{2}}][/tex]
n = 2
Thus, we can conclude that the final level of the electron is 2.
