Answer:
[tex]2.044 * 10^{-18}[/tex] J
Explanation:
Parameters given:
[tex]n_{1} = 4\\\\n_{2} = 1[/tex]
We know that the electron fell from level 4 to level 1. We can use the Rydberg's equation to find the [tex]2.044 * 10^{-18}[/tex] that is emitted by the electron in the process. Rydberg's equation is given as:
1/λ = [tex]R * (\frac{1}{(n_{2})^2} - \frac{1}{(n_{1})^2})[/tex]
where R = Rydberg's constant = [tex]1.0973731568508 * 10^{7}[/tex]
1/λ = [tex]1.0973731568508 * 10^{7} (\frac{1}{1^{2}} - \frac{1}{4^{2}} } )[/tex]
1/λ = [tex]1.0973731568508 * 10^{7} (\frac{1}{1} - \frac{1}{16} } )[/tex]
1/λ = [tex]1.0973731568508 * 10^{7} * \frac{15}{16}[/tex]
1/λ = [tex]1.029 * 10^{7}[/tex] [tex]m^{-1}[/tex]
∴ λ = [tex]9.72 * 10^{-8}[/tex] m
Energy of the photon is given by:
E = (hc)/λ
where
h = Planck's constant = [tex]6.62607004 * 10^{-34} m^2 kg / s[/tex]
c = speed of light = 299792458 m / s
∴ E = [tex]\frac{6.62607004 * 10^{-34} * 299792458}{9.72 * 10^{-8}}[/tex]
E = [tex]2.044 * 10^{-18}[/tex] J
The energy of the photon is [tex]2.044 * 10^{-18}[/tex] J
