When you irradiate a metal with light of wavelength 417 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.15 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts

Respuesta :

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.