Chapter 30, Problem 4TP

To determine

The energies of the first five excited states above the ground level for hydrogen atoms.

Answer to Problem 4TP

The energies of the first five excited states above the ground level for hydrogen atoms are $-3.396\text{eV}$, $-1.507\text{eV}$, $-0.848\text{eV}$, $-0.54\text{eV}$, and $-0.372\text{eV}$.

Explanation of Solution

Given:

The wavelength of first photon is ${\lambda }_1=121.6\text{nm}$.

The wavelength of second photon is ${\lambda }_2=102.6\text{nm}$.

The wavelength of third photon is ${\lambda }_3=97.3\text{nm}$.

The wavelength of fourth photon is ${\lambda }_4=95.0\text{nm}$.

The wavelength of five photon is ${\lambda }_5=93.8\text{nm}$.

Formula used:

The energy of the first excited state is given by

  $E_1=E_0+\frac{hc}{{\lambda }_1}$

The energy of the second excited state is given by

  $E_2=E_0+\frac{hc}{{\lambda }_2}$

The energy of the third excited state is given by

  $E_3=E_0+\frac{hc}{{\lambda }_3}$

The energy of the fourth excited state is given by

  $E_4=E_0+\frac{hc}{{\lambda }_4}$

The energy of the fifth excited state is given by

  $E_5=E_0+\frac{hc}{{\lambda }_5}$

Calculation:

The energy of the first excited state is calculated as follows:

  $\begin{array}{c}{E}_1=E_0+\frac{hc}{{\lambda }_1}\\ =-13.6\text{eV+}\frac{\left(4.1357\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3\times {10}^{17}\text{nm}/\text{s}\right)}{121.6\text{nm}}\\ =-13.6\text{eV}+\frac{12.407\times {10}^2\text{eV}\cdot \text{nm}}{121.6\text{nm}}\\ =-3.396\text{eV}\end{array}$

The energy of the second excited state is calculated as follows:

  $\begin{array}{c}{E}_2=E_0+\frac{hc}{{\lambda }_2}\\ =-13.6\text{eV+}\frac{\left(4.1357\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3\times {10}^{17}\text{nm}/\text{s}\right)}{102.6\text{nm}}\\ =-13.6\text{eV}+\frac{12.407\times {10}^2\text{eV}\cdot \text{nm}}{102.6\text{nm}}\\ =-1.507\text{eV}\end{array}$

The energy of the third excited state is calculated as follows:

  $\begin{array}{c}{E}_3=E_0+\frac{hc}{{\lambda }_3}\\ =-13.6\text{eV+}\frac{\left(4.1357\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3\times {10}^{17}\text{nm}/\text{s}\right)}{97.3\text{nm}}\\ =-13.6\text{eV}+\frac{12.407\times {10}^2\text{eV}\cdot \text{nm}}{97.3\text{nm}}\\ =-0.848\text{eV}\end{array}$

The energy of the fourth excited state is calculated as follows:

  $\begin{array}{c}{E}_4=E_0+\frac{hc}{{\lambda }_4}\\ =-13.6\text{eV+}\frac{\left(4.1357\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3\times {10}^{17}\text{nm}/\text{s}\right)}{95.0\text{nm}}\\ =-13.6\text{eV}+\frac{12.407\times {10}^2\text{eV}\cdot \text{nm}}{95.0\text{nm}}\\ =-0.54\text{eV}\end{array}$

The energy of the fifth excited state is given by

  $\begin{array}{c}{E}_5=E_0+\frac{hc}{{\lambda }_5}\\ =-13.6\text{eV+}\frac{\left(4.1357\times {10}^{-15}\text{eV}\cdot \text{s}\right)\left(3\times {10}^{17}\text{nm}/\text{s}\right)}{93.8\text{nm}}\\ =-13.6\text{eV}+\frac{12.407\times {10}^2\text{eV}\cdot \text{nm}}{93.8\text{nm}}\\ =-0.372\text{eV}\end{array}$

Conclusion:

The energies of the first five excited states above the ground level for hydrogen atoms are $-3.396\text{eV}$, $-1.507\text{eV}$, $-0.848\text{eV}$, $-0.54\text{eV}$, and $-0.372\text{eV}$.