Chapter 5, Problem 108A

Interpretation Introduction

Interpretation:

The wavelength of light capable of removing an electron from the gold surface with $8.17\times {10}^{-19}\text{J}$ of energy needs to be determined.

Concept introduction:

The relation between energy of liquid, wavelength, frequency and speed of light is represented as follows:

${\text{E}}_{\text{photon}}=\text{h}v$

and

$\text{λ}=\frac{c}{v}$

Here, E represents the energy of the light, h represents the Plank’s constant, c is the speed of the light, →is the wavelength of light and →is the frequency of light.

Answer to Problem 108A

Wavelength of the light is $2.43\times {10}^{-7}\text{m}$.

Explanation of Solution

Frequency can be calculated by using the equation given below where E represents the energy of the photon i.e. $8.17\times {10}^{-19}\text{J}$ and h represents the Plank’s constant i.e. $6.626\times {10}^{-34}\text{J}\text{.s}$.

$\begin{array}{c}{\text{E}}_{\text{photon}}=\text{h}v\\ v=\frac{{\text{E}}_{\text{photon}}}{\text{h}}\\ =\frac{8.17\times {10}^{-19}\text{J}}{6.626\times {10}^{-34}\text{J}\text{.s}}\\ =1.23\times {10}^{15}{\text{s}}^{-1}\end{array}$

Hence, the frequency of the light is $1.23\times {10}^{15}{\text{s}}^{-1}$.

Wavelength of the light can be determined by dividing the speed of the light by frequency. Speed of the light is $3.00\times {10}^8\text{m/s}$ and frequency is $1.23\times {10}^{15}{\text{s}}^{-1}$.

$\begin{array}{c}\text{λ}=\frac{c}{v}\\ =\frac{3.00\times {10}^8\text{m/s}}{1.23\times {10}^{15}{\text{s}}^{-1}}\\ =2.43\times {10}^{-7}\text{m}\end{array}$

Hence, the wavelength of the light is $2.43\times {10}^{-7}\text{m}$.

Conclusion

Wavelength of the light is $2.43\times {10}^{-7}\text{m}$.