Chapter 6.3, Problem 1PPA

Interpretation Introduction

Interpretation:

The energy of aphoton with a given wavelength and energy of the photon with a given frequency is to be calculated. The kinetic energy of anelectron ejected by the photon with wavelength $\lambda =2.11\times {10}^2\text{ nm}$ from a metal with given binding energy is to be calculated.

Concept introduction:

A photon is a dimensionless, massless, and charge-less quantum unit of space (particle)with an oscillating electromagnetic field (waves) that moves in space with constant speed (c) and finite energy (E).

The relation between the speed of light, wavelength, and frequency is as follows:

$c=v\lambda$

Here, $v$ is the frequency, $\lambda$ is the wavelength, and $c$ is the speed of light.

The frequency of electromagnetic radiation is used to determine the amount of energy carried by one of its photons.

The energy of the photon is directly proportional to frequency.

$E=hv$

Here, $v$ is the frequency, $E$ is the energy, and $h$ is Plank’s constant ($\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{ J s}$).

The relation between kinetic energy and binding energy is as follows:

$\text{KE}=hv-W$

Here, $\text{KE}$ is the kinetic energy of the ejected electron and $W$ is the binding energy of the electron.

The relation between $\text{eV}$ and joules is:

$1\text{eV}=1.6\times {10}^{-19}\text{J}$

The conversion factor to obtain the units of energy in $\text{J}$ is $\frac{1.6\times {10}^{-19}\text{J}}{1\text{ eV}}$.