Respuesta :
Answer:
The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is [tex]5.23\times 10^{-5} m[/tex]
Explanation:
Given :
The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:
[tex]E=\frac{R_y}{n^2}[/tex]
[tex]R_y=2.18\times 10^{-18} J[/tex] = Rydberg energy
n = principal quantum number of the orbital
Energy of 11th orbit = [tex]E_{11}[/tex]
[tex]E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J[/tex]
Energy of 10th orbit = [tex]E_{10}[/tex]
[tex]E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J[/tex]
Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.
[tex]E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J[/tex]
[tex]=E'=0.38\times 10^{-20} J[/tex]
[tex]\lambda =\frac{hc}{E'}[/tex] (Planck's' equation)
[tex]\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}[/tex]
[tex]\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m[/tex]
The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is [tex]5.23\times 10^{-5} m[/tex]
