Chapter 9, Problem 9.13E

The following are the numbers $n_2$ for some of the series of lines in the hydrogen atom spectrum:

Lyman $:1$ Balmer $:2$ Paschen $:3$ Brackett $:4$ Pfund $:5$

Calculate the energy changes, in ${\text{cm}}^{-1}$, of the lines in each of the stated series for each of the given values for $n_1$: (a) Lyman, $n_1=5$; (b) Balmer, $n_1=8$; (c) Paschen, $n_1=4$; (d) Brackett, $n_1=8$; (e) Pfund, $n_1=6$.

Interpretation Introduction

(a)

Interpretation:

The energy change in ${\text{cm}}^{-1}$, of the line in the given series for the given values for $n_1$ is to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Answer:

The energy change in ${\text{cm}}^{-1}$ for the Lyman series is $105347.52{\text{cm}}^{-1}$.

Explanation:

The final energy level for the Lyman series is $1$.

The given initial energy level for the Lyman series is $5$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Lyman series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(5\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(1-0.04\right)\\ & =& 105347.52{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Lyman series is $105347.52{\text{cm}}^{-1}$.

Conclusion:

The energy change in ${\text{cm}}^{-1}$ for the Lyman series is $105347.52{\text{cm}}^{-1}$.

Answer to Problem 9.13E

The energy change in ${\text{cm}}^{-1}$ for the Lyman series is $105347.52{\text{cm}}^{-1}$.

Explanation of Solution

The final energy level for the Lyman series is $1$.

The given initial energy level for the Lyman series is $5$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Lyman series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(5\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(1-0.04\right)\\ & =& 105347.52{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Lyman series is $105347.52{\text{cm}}^{-1}$.

Conclusion

The energy change in ${\text{cm}}^{-1}$ for the Lyman series is $105347.52{\text{cm}}^{-1}$.

Interpretation Introduction

(b)

Interpretation:

The energy change in ${\text{cm}}^{-1}$, of the line in the given series for the given values for $n_1$ is to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Answer:

The energy change in ${\text{cm}}^{-1}$ for the Balmer series is $25719.61{\text{cm}}^{-1}$.

Explanation:

The final energy level for the Balmer series is $2$.

The given initial energy level for the Balmer series is $8$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Balmer series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(8\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(0.25-0.015625\right)\\ & =& 25719.61{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Balmer series is $25719.61{\text{cm}}^{-1}$.

Conclusion:

The energy change in ${\text{cm}}^{-1}$ for the Balmer series is $25719.61{\text{cm}}^{-1}$.

Answer to Problem 9.13E

The energy change in ${\text{cm}}^{-1}$ for the Balmer series is $25719.61{\text{cm}}^{-1}$.

Explanation of Solution

The final energy level for the Balmer series is $2$.

The given initial energy level for the Balmer series is $8$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Balmer series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(8\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(0.25-0.015625\right)\\ & =& 25719.61{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Balmer series is $25719.61{\text{cm}}^{-1}$.

Conclusion

The energy change in ${\text{cm}}^{-1}$ for the Balmer series is $25719.61{\text{cm}}^{-1}$.

Interpretation Introduction

(c)

Interpretation:

The energy change in ${\text{cm}}^{-1}$, of the line in the given series for the given values for $n_1$ is to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Answer:

The energy change in ${\text{cm}}^{-1}$ for the Paschen series is $533.21{\text{cm}}^{-1}$.

Explanation:

The final energy level for the Paschen series is $3$.

The given initial energy level for the Paschen series is $4$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Paschen series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(3\right)}^2}-\frac{1}{{\left(4\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(0.1111-0.0625\right)\\ & =& 533.21{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Paschen series is $533.21{\text{cm}}^{-1}$.

Conclusion:

The energy change in ${\text{cm}}^{-1}$ for the Paschen series is $533.21{\text{cm}}^{-1}$.

Answer to Problem 9.13E

The energy change in ${\text{cm}}^{-1}$ for the Paschen series is $533.21{\text{cm}}^{-1}$.

Explanation of Solution

The final energy level for the Paschen series is $3$.

The given initial energy level for the Paschen series is $4$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Paschen series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(3\right)}^2}-\frac{1}{{\left(4\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(0.1111-0.0625\right)\\ & =& 533.21{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Paschen series is $533.21{\text{cm}}^{-1}$.

Conclusion

The energy change in ${\text{cm}}^{-1}$ for the Paschen series is $533.21{\text{cm}}^{-1}$.

Interpretation Introduction

(d)

Interpretation:

The energy change in ${\text{cm}}^{-1}$, of the line in the given series for the given values for $n_1$ is to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Answer:

The energy change in ${\text{cm}}^{-1}$ for the Brackett series is $5143.92{\text{cm}}^{-1}$.

Explanation:

The final energy level for the Brackett series is $4$.

The given initial energy level for the Brackett series is $8$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Brackett series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(4\right)}^2}-\frac{1}{{\left(8\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(0.0625-0.015625\right)\\ & =& 5143.92{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Brackett series is $5143.92{\text{cm}}^{-1}$.

Conclusion:

The energy change in ${\text{cm}}^{-1}$ for the Brackett series is $5143.92{\text{cm}}^{-1}$.

Answer to Problem 9.13E

The energy change in ${\text{cm}}^{-1}$ for the Brackett series is $5143.92{\text{cm}}^{-1}$.

Explanation of Solution

The final energy level for the Brackett series is $4$.

The given initial energy level for the Brackett series is $8$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Brackett series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(4\right)}^2}-\frac{1}{{\left(8\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(0.0625-0.015625\right)\\ & =& 5143.92{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Brackett series is $5143.92{\text{cm}}^{-1}$.

Conclusion

The energy change in ${\text{cm}}^{-1}$ for the Brackett series is $5143.92{\text{cm}}^{-1}$.

Interpretation Introduction

(e)

Interpretation:

The energy change in ${\text{cm}}^{-1}$, of the line in the given series for the given values for $n_1$ is to be calculated.

Concept introduction:

Rydberg equation is used to represent the wavenumber or wavelength of the lines present in the atomic spectrum of an element. The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Answer:

The energy change in ${\text{cm}}^{-1}$ for the Pfund series is $1338.79{\text{cm}}^{-1}$.

Explanation:

The final energy level for the Pfund series is $5$.

The given initial energy level for the Pfund series is $6$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Pfund series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(5\right)}^2}-\frac{1}{{\left(6\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(0.04-0.0278\right)\\ & =& 1338.79{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Pfund series is $1338.79{\text{cm}}^{-1}$.

Conclusion:

The energy change in ${\text{cm}}^{-1}$ for the Pfund series is $1338.79{\text{cm}}^{-1}$.

Answer to Problem 9.13E

The energy change in ${\text{cm}}^{-1}$ for the Pfund series is $1338.79{\text{cm}}^{-1}$.

Explanation of Solution

The final energy level for the Pfund series is $5$.

The given initial energy level for the Pfund series is $6$.

The Rydberg equation for the hydrogen atom is represented as,

$\overline{\nu }=R_{\text{H}}\left(\frac{1}{n_{\text{f}}^2}-\frac{1}{n_{\text{i}}^2}\right)$

Where,

$R_{\text{H}}$ represents Rydberg constant with a value for hydrogen $1.09737\times {10}^7{\text{m}}^{-1}$.

$n_{\text{i}}$ represents the initial energy level.

$n_{\text{f}}$ represents the final energy level.

Substitute the value of $R_{\text{H}}$, $n_{\text{i}}$, and $n_{\text{f}}$ for Pfund series in the above equation.

$\begin{array}{rll}\overline{\nu }& =& \left(1.09737\times {10}^7{\text{m}}^{-1}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{{\left(5\right)}^2}-\frac{1}{{\left(6\right)}^2}\right)\\ & =& \left(1.09737\times {10}^5{\text{cm}}^{-1}\right)\left(0.04-0.0278\right)\\ & =& 1338.79{\text{cm}}^{-1}\end{array}$

Therefore, the energy change in ${\text{cm}}^{-1}$ for the Pfund series is $1338.79{\text{cm}}^{-1}$.

Conclusion

The energy change in ${\text{cm}}^{-1}$ for the Pfund series is $1338.79{\text{cm}}^{-1}$.