Question

Photoelectrons with a maximum speed of 7.0 x 10^5 m/s are ejected from a surface in the presence of light with a frequency of 8.00 x 10^14 Hz. Calculate the energy required to eject photoelectrons from the surface

Answer 1

The energy required to eject photoelectrons from the surface is
`53 * 10^(-24) J`

Step-by-step explanation:

Given:

Speed of the photo electron=
`7.0 * 10^(5) \mathrm{m} / \mathrm{s}`

Frequency of the photoelectron=
`8.00 * 10^(14) \mathrm{Hz}`

To Find:

The energy required to eject photoelectrons from the surface.

Solution:

Energy of the photon is given by

E = hf

Where,

E = energy of the photons

h = Planck’s constant and its value is
`6.626 * 10^(-34) \mathrm{Js}`

Substituting the values in the formula we get ,

E = hf

`E=6.626 * 10^(-34) * 8 * 10^(14)`

`E=53 * 10^(-24) J`

Result:

Thus a energy of
`E=53 * 10^(-24) J` is required to eject thephotoelectrons from the surface