Answer:
A photon of wavelength 103 nm is released
Explanation:
When an electron in an atom jumps from a higher energy level to a lower energy level, it releases a photon whose energy is equal to the difference in energy between the two levels.
For example, if we are talking about a hydrogen atom, the energy of the levels are:
[tex]E_1 = -13.6 eV\\E_2 = -3.4 eV\\E_3 = -1.5 eV[/tex]
So, the energy of the photon released when the electron jumps from the level n=3 to n=1 is
[tex]\Delta E = E_3 - E_1 = -1.5 -(-13.6)=12.1 eV[/tex]
In Joules,
[tex]\Delta E =12.1\cdot 1.6\cdot 10^{-19} = 1.94\cdot 10^{-18} J[/tex]
We can also find the wavelength of this photon, using the equation:
[tex]\Delta E = \frac{hc}{\lambda}\rightarrow \lambda=\frac{hc}{\Delta E}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.94\cdot 10^{-18}}=1.03\cdot 10^{-7} m = 103 nm[/tex]