Chapter 5, Problem 119A

Interpretation Introduction

Interpretation:

The energy of the photons related to the two bright lines 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 119A

Energy for the first line is $3.38\times {10}^{-19}\text{J}$ and energy for the second line is $3.37\times {10}^{-19}\text{J}$.

Explanation of Solution

Given Information:

The wavelength for the first line is $5.89\times {10}^{-7}\text{m}$ and the wavelength for the second line is $5.90\times {10}^{-7}\text{m}$.

Calculation:

To determine the energies of the two lines, use the wavelength of the both lines. The wavelength for the first line is $5.89\times {10}^{-7}\text{m}$ and the wavelength for the second line is $5.90\times {10}^{-7}\text{m}$. The value of h represents the Plank’s constant i.e. $6.626\times {10}^{-34}\text{J}\text{.s}$ and c represents the speed of the light is $3.00\times {10}^8\text{m/s}$.

$E=\frac{hc}{\text{λ}}$

Calculation of energy for the first line as:

$\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\times \left(3.00\times {10}^8\text{m/s}\right)}{\left(5.89\times {10}^{-7}\text{m}\right)}\\ =3.38\times {10}^{-19}\text{J}\end{array}$

Calculation of energy for the second line as:

$\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\times \left(3.00\times {10}^8\text{m/s}\right)}{\left(5.90\times {10}^{-7}\text{m}\right)}\\ =3.37\times {10}^{-19}\text{J}\end{array}$

Hence, the energy for the first line is $3.38\times {10}^{-19}\text{J}$ and energy for the second line is $3.37\times {10}^{-19}\text{J}$.

Conclusion

Energy for the first line is $3.38\times {10}^{-19}\text{J}$ and energy for the second line is $3.37\times {10}^{-19}\text{J}$.