Chapter 9, Problem 9.40E

Interpretation Introduction

Interpretation:

The energy of a single photon in joules and the energy of a mole of photons in $\text{J}/\text{mol}$ for light having the given wavelengths are to be calculated. Whether the calculated energies explain the relative danger of electromagnetic radiation of differing wavelengths or not is to be stated.

Concept introduction:

A small packet of energy is known as the quanta. Light is emitted in the form of quanta or photons. The Planck’s law gives the relation between the energy and wavelength, frequency and wavenumber.

Answer to Problem 9.40E

The energy of single photon in joules and the energy of a mole of photons in $\text{J}/\text{mol}$ for light having wavelength of $\text{10}\text{m}$ (radio and TV waves) is $1.99\times {10}^{-26}\text{J}$ and $1.2\times {10}^{-2}\text{J}/\text{mol}$, $\text{10}\text{.0}\text{cm}$ (microwaves) is $1.99\times {10}^{-23}\text{J}$ and $1.2\text{J}/\text{mol}$, $\text{10}\text{microns}$ (infrared range) is $1.99\times {10}^{-20}\text{J}$ and $1.2\times {10}^4\text{J}/\text{mol}$, $\text{550}\text{nm}$ (green light) is $3.6\times {10}^{-19}\text{J}$ and $2.4\times {10}^5\text{J}/\text{mol}$, $\text{300}\text{nm}$ (ultraviolet) is $6.6\times {10}^{-19}\text{J}$ and $4.0\times {10}^5\text{J}/\text{mol}$ and $\text{1}\text{.00}\stackrel{\circ }{\text{A}}$ (X rays) is $1.99\times {10}^{-15}\text{J}$ and $1.2\times {10}^9\text{J}/\text{mol}$.

The shorter wavelength radiations are more dangerous because these radiations transmit more energy.

Explanation of Solution

The energy of single photon in joules is calculated by the formula,

$E=h\nu$ The formula can be written as follows:

$E=h\frac{c}{\lambda }$…$\left(1\right)$

Where,

• $h$ is the planks constant.

• $c$ is the velocity of light.

• $\lambda$ is the wave length.

The energy of a mole of photons in $\text{J}/\text{mol}$ is calculated by the formula,

$E=\frac{N_{\text{A}}hc}{\lambda }$ …$\left(2\right)$

Where,

• $N_{\text{A}}$ is the Avogadro number.

• $E$ is the energy of a mole of photon in $\text{J}/\text{mol}$.

The given wavelength of light is $\text{10}\text{m}$ (radio and TV waves).

Substitute the value of $\lambda$ and $c$ in the equation $\left(1\right)$

$\begin{array}{rll}E& =& \frac{hc}{\lambda }\\ & =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{10}\text{m}}\\ & =& 1.99\times {10}^{-26}\text{J}\end{array}$

Thus, the energy of single photon in joules is $1.99\times {10}^{-26}\text{J}$.

Substitute the value of $\lambda$ and $c$ in the equation $\left(2\right)$.

$\begin{array}{rll}E& =& \left(6.022\times {10}^{23}{\text{mol}}^{-1}\right)\times \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{10}\text{m}}\\ E& =& 1.2\times {10}^{-2}\text{J}/\text{mol}\end{array}$

Thus, the energy of a mole of photons in $\text{J}/\text{mol}$ is $1.2\times {10}^{-2}\text{J}/\text{mol}$.

The given wavelength of light is $\text{10}\text{.0}\text{cm}$ (microwaves).

Substitute the value of $\lambda$ and $c$ in the equation $\left(1\right)$

$\begin{array}{rll}E& =& \frac{hc}{\lambda }\\ & =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{10}\text{.0}\text{cm}\times \left(\frac{{10}^{-2}\text{m}}{\text{10}\text{.0}\text{cm}}\right)}\\ & =& 1.99\times {10}^{-23}\text{J}\end{array}$

Thus, the energy of single photon in joules is $1.99\times {10}^{-24}\text{J}$.

Substitute the value of $\lambda$ and $c$ in the equation $\left(2\right)$.

$\begin{array}{rll}E& =& \left(6.022\times {10}^{23}{\text{mol}}^{-1}\right)\times \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{10}\text{.0}\text{cm}\times \left(\frac{{10}^{-2}\text{m}}{\text{10}\text{.0}\text{cm}}\right)}\\ E& =& 1.2\text{J}/\text{mol}\end{array}$

Thus, the energy of a mole of photons in $\text{J}/\text{mol}$ is $1.2\text{J}/\text{mol}$.

The given wavelength of light is $\text{10}\text{microns}$ (infrared range).

Substitute the value of $\lambda$ and $c$ in the equation $\left(1\right)$

$\begin{array}{rll}E& =& \frac{hc}{\lambda }\\ & =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{10}\text{microns}\times \left(\frac{{10}^{-6}\text{m}}{\text{1}\text{microns}}\right)}\\ & =& 1.99\times {10}^{-20}\text{J}\end{array}$

Thus, the energy of single photon in joules is $1.99\times {10}^{-20}\text{J}$.

Substitute the value of $\lambda$ and $c$ in the equation $\left(2\right)$.

$\begin{array}{rll}E& =& \left(6.022\times {10}^{23}{\text{mol}}^{-1}\right)\times \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{10}\text{microns}\times \left(\frac{{10}^{-6}\text{m}}{\text{1}\text{microns}}\right)}\\ E& =& 1.2\times {10}^4\text{J}/\text{mol}\end{array}$

Thus, the energy of a mole of photons in $\text{J}/\text{mol}$ is $1.2\times {10}^4\text{J}/\text{mol}$.

The given wavelength of light is $\text{550}\text{nm}$ (green light).

Substitute the value of $\lambda$ and $c$ in the equation $\left(1\right)$

$\begin{array}{rll}E& =& \frac{hc}{\lambda }\\ & =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{550.0\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ & =& 3.6\times {10}^{-19}\text{J}\end{array}$

Thus, the energy of single photon in joules is $3.6\times {10}^{-19}\text{J}$.

Substitute the value of $\lambda$ and $c$ in the equation $\left(2\right)$.

$\begin{array}{rll}E& =& \left(6.022\times {10}^{23}{\text{mol}}^{-1}\right)\times \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{550.0\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ E& =& 2.4\times {10}^5\text{J}/\text{mol}\end{array}$

Thus, the energy of a mole of photons in $\text{J}/\text{mol}$ is $2.4\times {10}^5\text{J}/\text{mol}$.

The given wavelength of light is $\text{300}\text{nm}$ (ultraviolet).

Substitute the value of $\lambda$ and $c$ in the equation $\left(1\right)$

$\begin{array}{rll}E& =& \frac{hc}{\lambda }\\ & =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{300}\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ & =& 6.6\times {10}^{-19}\text{J}\end{array}$

Thus, the energy of single photon in joules is $6.6\times {10}^{-19}\text{J}$.

Substitute the value of $\lambda$ and $c$ in the equation $\left(2\right)$.

$\begin{array}{rll}E& =& \left(6.022\times {10}^{23}{\text{mol}}^{-1}\right)\times \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{300}\text{nm}\times \left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ E& =& 4.0\times {10}^5\text{J}/\text{mol}\end{array}$

Thus, the energy of a mole of photons in $\text{J}/\text{mol}$ is $4.0\times {10}^5\text{J}/\text{mol}$.

The given wavelength of light is $\text{1}\text{.00}\stackrel{\circ }{\text{A}}$ (X rays).

Substitute the value of $\lambda$ and $c$ in the equation $\left(1\right)$

$\begin{array}{rll}E& =& \frac{hc}{\lambda }\\ & =& \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{1}\text{.00}\stackrel{°}{\text{A}}\times \left(\frac{{10}^{-10}\text{m}}{\text{1}\text{.00}\stackrel{°}{\text{A}}}\right)}\\ & =& 1.99\times {10}^{-15}\text{J}\end{array}$

Thus, the energy of single photon in joules is $1.99\times {10}^{-15}\text{J}$.

Substitute the value of $\lambda$ and $c$ in the equation $\left(2\right)$.

$\begin{array}{rll}E& =& \left(6.022\times {10}^{23}{\text{mol}}^{-1}\right)\times \frac{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.0\times {10}^8\text{m}{\text{s}}^{-1}\right)}{\text{1}\text{.00}\stackrel{°}{\text{A}}\times \left(\frac{{10}^{-10}\text{m}}{\text{1}\text{.00}\stackrel{°}{\text{A}}}\right)}\\ E& =& 1.2\times {10}^9\text{J}/\text{mol}\end{array}$

Thus, the energy of a mole of photons in $\text{J}/\text{mol}$ is $1.2\times {10}^9\text{J}/\text{mol}$.

It is observed from the above calculated values that the energy of per mole for shorter wavelength radiations is higher than the energy of per mole for longer wavelength radiations. Thus, the shorter wavelength radiations are more dangerous.

Conclusion

The energy of single photon in joules and the energy of a mole of photons in $\text{J}/\text{mol}$ for light having wavelength of $\text{10}\text{m}$ (radio and TV waves) is $1.99\times {10}^{-26}\text{J}$ and $1.2\times {10}^{-2}\text{J}/\text{mol}$, $\text{10}\text{.0}\text{cm}$ (microwaves) is $1.99\times {10}^{-23}\text{J}$ and $1.2\text{J}/\text{mol}$, $\text{10}\text{microns}$ (infrared range) is $1.99\times {10}^{-20}\text{J}$ and $1.2\times {10}^4\text{J}/\text{mol}$, $\text{550}\text{nm}$ (green light) is $3.6\times {10}^{-19}\text{J}$ and $2.4\times {10}^5\text{J}/\text{mol}$, $\text{300}\text{nm}$ (ultraviolet) is $6.6\times {10}^{-19}\text{J}$ and $4.0\times {10}^5\text{J}/\text{mol}$ and $\text{1}\text{.00}\stackrel{\circ }{\text{A}}$ (X rays) is $1.99\times {10}^{-15}\text{J}$ and $1.2\times {10}^9\text{J}/\text{mol}$.

The shorter wavelength radiations are more dangerous because these radiations transmit more energy.