Chapter 2.9, Problem 36AAP

To determine

The number of photons needed if UV light is used.

Answer to Problem 36AAP

The number of photons needed if UV light is used is $\text{1}\text{.68}\times {\text{10}}^{21}\text{photons}$.

The number of photons needed if visible light is used is $\text{8}\text{.36}\times {\text{10}}^{22}\text{photons}$.

The number of photons needed if infrared light is used is $\text{1}\text{.6}\times {\text{10}}^{25}\text{photons}$.

The energy of the UV light is greater when compared to other electromagnetic radiations.

 $E_{\text{UV}}>E_{\text{visible}}>E_{\text{infrared}}$

Explanation of Solution

Write the expression to calculate energy of the electromagnetic radiation $\left(E\right)$.

 $E=\frac{hc}{\lambda }$                                                                                                             (I)

Here, Planck's constant is $h$, speed of light is c and wavelength of the light is $\lambda$.

Write the expression to calculate number of photons needed $\left(n\right)$.

 $n=\frac{E_{\text{in}}}{E}$                                                                                                            (II)

Here, input energy is $E_{\text{in}}$.

Conclusion:

The values of Planck constant and speed of light are $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ and $3\times {10}^8\text{m}/\text{s}$ respectively.

For UV light:

It is given that the wavelength of UV light is $\text{1}\times {\text{10}}^{-8}\text{m}$.

 $\lambda =\text{1}\times {\text{10}}^{-8}\text{m}$

Substitute $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for c and $\text{1}\times {\text{10}}^{-8}\text{m}$ for $\lambda$ in Equation (I).

 $\begin{array}{c}E=\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\text{1}\times {\text{10}}^{-8}\text{m}}\\ =1.989\times {10}^{-17}\text{J}\end{array}$

Substitute $33400\text{J}$ for $E_{\text{in}}$ and $1.989\times {10}^{-17}\text{J}$ for $E$ in Equation (II).

 $\begin{array}{c}n=\frac{33400\text{J}}{1.989\times {10}^{-17}\text{J}}\\ =\text{1}\text{.68}\times {\text{10}}^{21}\text{photons}\end{array}$

Thus, the number of photons needed if UV light is used is $\text{1}\text{.68}\times {\text{10}}^{21}\text{photons}$.

For visible light:

It is given that the wavelength of visible light is $\text{5}\times {\text{10}}^{-7}\text{m}$.

 $\lambda =\text{5}\times {\text{10}}^{-7}\text{m}$

Substitute $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for c and $\text{5}\times {\text{10}}^{-7}\text{m}$ for $\lambda$ in Equation (I).

 $\begin{array}{c}E=\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\text{5}\times {\text{10}}^{-7}\text{m}}\\ =4\times {10}^{-19}\text{J}\end{array}$

Substitute $33400\text{J}$ for $E_{\text{in}}$ and $4\times {10}^{-19}\text{J}$ for $E$ in Equation (II).

 $\begin{array}{c}n=\frac{33400\text{J}}{4\times {10}^{-19}\text{J}}\\ =\text{8}\text{.36}\times {\text{10}}^{22}\text{photons}\end{array}$

Thus, the number of photons needed if visible light is used is $\text{8}\text{.36}\times {\text{10}}^{22}\text{photons}$.

For infrared light:

It is given that the wavelength of infrared light is $1\times {\text{10}}^{-4}\text{m}$.

 $\lambda =1\times {\text{10}}^{-4}\text{m}$

Substitute $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for c and $1\times {\text{10}}^{-4}\text{m}$ for $\lambda$ in Equation (I).

 $\begin{array}{c}E=\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{1\times {\text{10}}^{-4}\text{m}}\\ =2\times {10}^{-21}\text{J}\end{array}$

Substitute $33400\text{J}$ for $E_{\text{in}}$ and $2\times {10}^{-21}\text{J}$ for $E$ in Equation (II).

 $\begin{array}{c}n=\frac{33400\text{J}}{2\times {10}^{-21}\text{J}}\\ =\text{1}\text{.6}\times {\text{10}}^{25}\text{photons}\end{array}$

Thus, the number of photons needed if infrared light is used is $\text{1}\text{.6}\times {\text{10}}^{25}\text{photons}$.

The important conclusion is that the energy of the UV light is greater when compared to other electromagnetic radiations.

 $E_{\text{UV}}>E_{\text{visible}}>E_{\text{infrared}}$