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

Answer:

The energy required to eject photoelectrons from the surface is$53 * 10^{-24} J$

Explanation:

Given:

Speed of the photo electron=$7.0 \times 10^{5} \mathrm{m} / \mathrm{s}$

Frequency of the photoelectron= $8.00 \times 10^{14} \mathrm{Hz}$

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