The frequency of this photon is 1.20×[tex]10^1^5[/tex] Hz.
It is given that the photon in a laboratory experiment has an energy of 5 eV.
It is required to find the frequency of the photon.
It is defined as the number of waves that crosses a fixed point in one second known as frequency. The unit of frequency is per second.
We have the energy of the photon = 5 eV
We know the 1 eV = [tex]1.6\times10^{-19}[/tex] Joules.
5 eV = 5×[tex]1.6\times10^{-19}[/tex] Joules ⇒ [tex]8\times10^{-19}[/tex] Joules
Photon energy(E) is the multiplication of plank's constant (h) and frequency(v) ie.
E = hv
[tex]8\times10^{-19}[/tex] = [tex]6.63\times10^{-34}[/tex] ×v (h = [tex]6.63\times10^{-34}[/tex] [tex]\rm m^2 \ kg/s[/tex])
[tex]\rm v = \frac{8\times10^{-19}}{6.63\times10^{-34}}[/tex]
v = 1.20×[tex]10^1^5[/tex] Hz
Thus, the frequency of this photon is 1.20×[tex]10^1^5[/tex] Hz.
Learn more about the frequency here:
brainly.com/question/27063800
