We will find that the change in energy is 0.967 eV and the wavelength is 128nm
We know that the energy of the n-th state of a hydrogen atom is:
[tex]E_n = \frac{-13.6 eV}{n^2}[/tex]
Then when we transition from n = 3 to n = 5, the change in energy is given by:
[tex]E_5 - E_3 = \frac{-13.6 eV}{5^2} - \frac{-13.6 eV}{3^2} = 0.967 eV[/tex]
The frequency is given by the difference in energy (which we found above) divided by Planck's constant, so we will get:
[tex]\omega_{3->5} = \frac{E_5 - E_3}{h}[/tex]
Where the Planck's constant is:
h = 4.14*10^(-15) eV/Hz
Then we have:
[tex]\omega_{3->5} = \frac{0.967 eV}{4.14*10^{-15} eV/Hz} = 2.34*10^{14} Hz[/tex]
And the wavelength is given by the speed of the wave (speed of light) divided by the frequency, so we get:
[tex]\lambda = \frac{3*1^8 m/s}{2.34*10^{14} s^-1} = 1.28*10^{-6} m[/tex]
And we want this in nanometers, which is:
1nm = 1m*10^{-9}
Then:
[tex]\lambda = 1.28*10^{-6}m =1.28*10^{-6}*(10^9 nm) = 128 nm[/tex]
If you want to learn more about the hydrogen atom, you can read:
https://brainly.com/question/4597871
