Answer:
E = 4.76 eV
Explanation:
It is given that,
Wavelength = 417 nm
Potential difference = 1.15 V
We need to find the energy of a photon of this light in electron volts. The energy of a photon is given by :
[tex]E=\dfrac{hc}{\lambda}[/tex]
Where
h is Planck's constant, c is speed of light
[tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{417\times 10^{-9}}\\\\=4.76\times 10^{-19}\ J[/tex]
We know that, [tex]1\ eV=1.6\times 10^{-19}\ J[/tex]
or
E = 4.76 eV
So, the energy of a photon is 4.76 eV.