Answer:
2 eV
Explanation:
The energy of a photon of light is given by the formula
[tex]E=\frac{hc}{\lambda}[/tex]
where
h is the Planck constant
c is the speed of light
[tex]\lambda[/tex] is the wavelength of the photon
In this problem we have:
[tex]h=6.63\cdot 10^{-34} Js[/tex]
[tex]c=3.0\cdot 10^8 m/s[/tex]
[tex]\lambda=6250 A = 6250\cdot 10^{-10} m[/tex] is the wavelength of the photon
Therefore, the energy in Joules is
[tex]E=\frac{(6.63\cdot 10^{-34})(3.0\cdot 10^8)}{6250\cdot 10^{-10}}=3.2\cdot 10^{-19}J[/tex]
We want to convert this energy into electronvolts: we know that the conversion factor is
[tex]1 eV = 1.6\cdot 10^{-19}J[/tex]
Therefore,
[tex]E=\frac{3.2\cdot 10^{-19}}{1.6\cdot 10^{-19}}=2 eV[/tex]
