Chapter 5.3, Problem 18SP

Interpretation Introduction

Interpretation: For the given wavelength of light, the energy of the photon of light needs to be determined.

Concept Introduction:

The relation between wavelength and frequency of radiation is as follows:

  $v=\frac{c}{\lambda }$

Here, $v$ is the frequency of radiation, $\lambda$ is the wavelength of radiation, and c is the speed of light. The value of the speed of light is constant that is $3\times {10}^8\text{ m/s}$ .

Also, the energy of the photon is directly proportional to the frequency.

The relation is represented as follows:

  $E=hv$

Here, h is Planck’s constant $6.626\times {10}^{-34}\text{ Js}$

Answer to Problem 18SP

  $7.62\times {10}^{-19}\text{ J}$

Explanation of Solution

The given wavelength is 260 nm. From the given wavelength, the frequency can be calculated as follows:

  $v=\frac{c}{\lambda }$

Here, the value of the speed of light is constant that is $3\times {10}^8\text{ m/s}$ .

Substitute the values in the above formula,

  $\begin{array}{c}v=\frac{3\times {10}^8\text{ m/s}}{260\text{ nm}\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)}\\ =0.0115\times {10}^{17}{\text{ s}}^{-1}\\ =1.15\times {10}^{15}{\text{ s}}^{-1}\end{array}$

Now, the calculated frequency is $1.15\times {10}^{15}{\text{ s}}^{-1}$ . Energy of a photon of light be calculated using the following formula:

  $E=hv$

Substitute the values,

  $\begin{array}{c}E=\left(6.626\times {10}^{-34}\text{ Js}\right)\left(1.15\times {10}^{15}{\text{ s}}^{-1}\right)\\ =7.62\times {10}^{-34+15}\text{ J}\\ =7.62\times {10}^{-19}\text{ J}\end{array}$

Therefore, the energy of a photon of light is $7.62\times {10}^{-19}\text{ J}$ .

Conclusion

The energy of a photon of light is $7.62\times {10}^{-19}\text{ J}$ .