Chapter 43, Problem 9SP

To determine

The difference between the atom’s two energy levels of yellow sodium emission line, which has the wavelength of $587.5618\text{ nm}$.

Answer to Problem 9SP

Solution:

$2.110\text{ eV}$

Explanation of Solution

Given data:

The wavelength of the yellow sodium emission line is $587.5618\text{ nm}$.

Formula used:

The expression for the difference between the two energy levels is written as

$\Delta E=\frac{hc}{\lambda }$

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

Explanation:

Recall the expression for the difference between the two energy levels:

$\Delta E=\frac{hc}{\lambda }$

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$ and $587.5618\text{ nm}$ for $\lambda$

$\begin{array}{c}\Delta E=\frac{1240\text{ eV}\cdot \text{nm}}{587.5618\text{ nm}}\\ =2.110\text{ eV}\end{array}$

Conclusion:

Hence, the difference between the atom’s two energy levels of yellow sodium emission line is $2.110\text{ eV}$.