The energy of a photon is given by
[tex]E=hf[/tex]
where h is the Planck constant and f is the photon frequency.
We can find the photon's frequency by using the following relationship:
[tex]f= \frac{c}{\lambda} [/tex]
where c is the speed of light and [tex]\lambda[/tex] is the photon's wavelength. By plugging numbers into the equation, we find
[tex]f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.67 \cdot 10^{14}Hz [/tex]
And so now we can find the photon energy
[tex]E=hf=(6.6 \cdot 10^{-34} Js)(6.67 \cdot 10^{14}Hz )=4.4 \cdot 10^{-19} J[/tex]
We know that 1 Joule corresponds to
[tex]1 J = 1.6 \cdot 10^{-19} eV[/tex]
So we can convert the photon's energy into electronvolts:
[tex]E= \frac{4.4 \cdot 10^{-19} J }{1.6 \cdot 10^{-19} J/eV}=2.75 eV [/tex]
