Respuesta :
According to Bohr's Model of the Hydrogen Atom, the energy of the n-th state is given by the expression:
[tex]E_n=-13.6eV\left(\frac{1}{n^2}\right)[/tex]If an electron changes from a state n to a state m, the energy of the photon emitted is given by:
[tex]\Delta E=E_n-E_m=-13.6eV\left(\frac{1}{n}-\frac{1}{m}\right)[/tex]Replace n=6 and m=4 to find the energy of the emitted photon:
[tex]\Delta E=-13.6eV\left(\frac{1}{6}-\frac{1}{4}\right)=1.1333...eV[/tex]On the other hand, the energy of a photon and its wavelength are related through the equation:
[tex]E=\frac{hc}{\lambda}[/tex]Then, the wavelength of the photon emitted is:
[tex]\begin{gathered} \lambda=\frac{hc}{E} \\ \\ =\frac{(6.626\times10^{-34}Js)(2.9979\times10^8\frac{m}{s})}{1.1333...eV} \\ \\ =\frac{(6.626\times10^{-34}Js)(2.9979\times10^8\frac{m}{s})}{1.1333...\cdot1.602\times10^{-19}C\cdot V} \\ \\ =1.094...\times10^{-6}m \\ \\ \approx1094nm \end{gathered}[/tex]Therefore, the wavelength of the photon emitted is approximately 1094 nanometers.
