Chapter 27, Problem 11P

To determine

Energy range of photon in visible spectrum in $\text{J}$ and in $\text{eV}$

Answer to Problem 11P

Solution:

Energy range of photon in visible spectrum is $4.9\times {10}^{-19}\text{J}\sim 2.64\times {10}^{-19}\text{J}$ or $1.65\text{eV}\sim 3.0\text{eV}$.

Explanation of Solution

Given:

Wavelength range of visible spectrum, $\lambda =400\text{nm}\sim 750\text{nm}$

Formula used:

Energy is given by the formula

$E=\frac{hc}{\lambda }$

Here,

$h=6.62\times {10}^{-34}\text{J-s}$ is Planck’s constant

$\lambda$ is the wavelength

$c=3\times {10}^8\text{m/s}$ is the speed of light

Calculation:

The maximum energy in visible spectrum is calculated as

$\begin{array}{c}{E}_{\max }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}\\ =4.9\times {10}^{-19}\text{J}\end{array}$

The minimum energy in visible spectrum is calculated as

$\begin{array}{c}{E}_{\min }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{750\times {10}^{-9}}\\ =2.64\times {10}^{-19}\text{J}\end{array}$

Therefore the range of energy in visible spectrum is $4.9\times {10}^{-19}\text{J}\sim 2.64\times {10}^{-19}\text{J}$

The energy range in $\text{eV}$ is calculated as

$\begin{array}{c}{E}_{\max }=\frac{4.9\times {10}^{-19}}{1.6\times {10}^{-19}}\\ =3.0625\end{array}$

The minimum energy is calculated as

$\begin{array}{c}{E}_{\min }=\frac{2.64\times {10}^{-19}}{1.6\times {10}^{-19}}\\ =1.65\text{eV}\end{array}$

Therefore the range of energy in visible spectrum is $1.65\text{eV}\sim 3.0\text{eV}$