Question

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

Answer 1

E = 4.76 eV

Step-by-step 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 :

`E=(hc)/(\lambda)`

Where

h is Planck's constant, c is speed of light

`E=(6.63* 10^(-34)* 3* 10^8)/(417* 10^(-9))\\\\=4.76* 10^(-19)\ J`

We know that,
`1\ eV=1.6* 10^(-19)\ J`

or

E = 4.76 eV

So, the energy of a photon is 4.76 eV.