Chapter 6, Problem 54GQ

Excited H atoms have many emission lines. One series of lines, called the Pfund series, occurs in the infrared region. It results when an electron changes from higher energy levels to a level with n = 5. Calculate the wavelength and frequency of the lowest energy line of this series.

Interpretation Introduction

Interpretation: The wavelength and frequency of the least energetic in Pfund series of the excited $\text{H}$ atom is to be calculated.

Concept introduction:

• Electronic transitions that take place in excited $\text{H}$ atom is,

1. 1. 1. Lyman series: electronic transitions take place to the $\text{n}\text{=}\text{1}$ level and it is in ultraviolet region.
2. 2. Balmer series: electronic transitions take place from $\text{n}\text{>}\text{2}$ to the $\text{n}\text{=}2$ level and it is in visible region.
3. 3. Ritz-Paschen series: electronic transitions take place from $\text{n}\text{>}3$ to the $\text{n}\text{=}3$ level and it is in infrared region.
4. 4. Brackett series: electronic transitions take place from $\text{n}\text{>}4$ to the $\text{n}\text{=}4$ level.
5. 5. Pfund series: electronic transitions take place from $\text{n}\text{>}5$ to the $\text{n}\text{=}5$ level

$\begin{array}{c}\text{Energy}\text{between}\text{the}\text{states}\text{ΔE}\text{=}{\text{E}}_{\text{final}}-{\text{E}}_{\text{initial}}\\ \text{=}-\text{Rhc}\left(\frac{\text{1}}{{\text{n}}_{\text{final}}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{initial}}^{\text{2}}}\right)\\ \text{where,}\text{R}\text{=}\text{Rydberg}\text{constant}\\ \text{h}\text{=}\text{Planck's}\text{constant}\\ \text{c}\text{=}\text{speed}\text{of}\text{light}\\ \text{n}\text{=}\text{Principal}\text{quantum}\text{number}\end{array}$

As the energy gap between two transition states increases the wavelength of the radiation emitted decreases

• Planck’s equation,

    $\begin{array}{c}\text{E}\text{=}\text{hν}=\frac{\text{hc}}{\text{λ}}\\ \text{where, E}\text{=}\text{energy}\\ \text{h}\text{=}\text{Planck's}\text{constant}\\ \text{ν}=\text{frequency}\end{array}$

The energy increases as the wavelength of the light decrease. Also the energy increases as the frequency of the light increases.

• The frequency of the light is inversely proportional to its wavelength.

  $\begin{array}{c}\text{ν}\text{=}\frac{\text{c}}{\text{λ}}\\ \text{where, c}\text{=}\text{speed}\text{of}\text{light}\\ \text{ν}=\text{frequency}\\ \text{λ}=\text{wavelength}\end{array}$

Answer to Problem 54GQ

The wavelength of the least energetic in Pfund series of the excited $\text{H}$ atom is $7471.6\text{nm}$.

The frequency of the least energetic in Pfund series of the excited $\text{H}$ atom is $4.012\text{×}{\text{10}}^{13}{\text{s}}^{-1}$.

Explanation of Solution

As the energy gap between two transition states decreases the wavelength of the radiation emitted increases

Hence the line with highest wavelength is produced in Pfund series of $\text{H}$ atom when the electronic transitions that take place from the $\text{n}\text{=}6$ to the $\text{n}\text{=}5$.

Since $\text{E}\text{=}\frac{\text{hc}}{\text{λ}}$ ,the energy decreases as the wavelength of the light increases. The electronic transitions that taking place from the $\text{n}\text{=}6$ to the $\text{n}\text{=}5$ has the least energy

The lowest energy line in Pfund series of the excited $\text{H}$ atom forms if the transition of electron is from $\text{n}\text{=}6$ to the $\text{n}\text{=}5$

  $\begin{array}{c}\text{R}\text{=}\text{1}\text{.097}\text{×}{\text{10}}^{\text{7}}{\text{m}}^{-\text{1}}\\ \text{h}\text{=}\text{6}\text{.626}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s}\\ \text{c}\text{=}\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}\\ {\text{n}}_{\text{initial}}\text{=}6\\ {\text{n}}_{\text{final}}\text{=}5\end{array}$

Energy is determined,

$\begin{array}{c}\text{Energy}\text{between}\text{the}\text{states}\text{ΔE}\text{=}{\text{E}}_{\text{final}}-{\text{E}}_{\text{initial}}\\ \text{=}-\text{Rhc}\left(\frac{\text{1}}{{\text{n}}_{\text{final}}^{\text{2}}}-\frac{\text{1}}{{\text{n}}_{\text{initial}}^{\text{2}}}\right)\\ \text{=}-\text{1}\text{.097}\text{×}{\text{10}}^{\text{7}}{\text{m}}^{-\text{1}}\text{×}\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}\text{×}\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}\left(\frac{\text{1}}{5^{\text{2}}}-\frac{\text{1}}{6^{\text{2}}}\right)\\ \text{=}-\text{2}\text{.6634}\text{×}{\text{10}}^{-20}\text{J/photon}\end{array}$

The energy emitted by the photon does not have any sign and absolute value is taken. Hence, the value of energy is $\text{2}\text{.6634}\text{×}{\text{10}}^{-20}\text{J/photon}$

• The frequency of lowest energy line in Pfund series of the excited $\text{H}$ atom is calculated,

According to Planck’s equation,

  $\text{E}\text{=}\text{hν}$

Therefore,

  $\begin{array}{c}\text{ν}=\frac{\text{E}}{h}\\ \\ =\frac{\text{2}\text{.6634}\text{×}{\text{10}}^{-20}\text{J/photon}}{\text{6}\text{.626}\text{×}{\text{10}}^{-\text{34}}\text{J}\text{.s}/\text{photon}}\\ \\ =4.0196\text{×}{\text{10}}^{13}{\text{s}}^{-1}\end{array}$

The frequency of lowest energy line in Pfund series of the excited $\text{H}$ atom is $4.0196\times 10^{13}{s}^{-1}$

• The wavelength of lowest energy line in Pfund series of the excited $\text{H}$ atom is calculated,

  $\text{λ}\text{=}\frac{\text{c}}{\text{ν}}$

Substituting the values,

  $\begin{array}{c}\text{λ}=\frac{\text{2}\text{.998}\text{×}\text{10}{}^{\text{8}}\text{m/s}}{4.0196\text{×}{\text{10}}^{13}{\text{s}}^{-1}}\\ \\ =7.4584\times {10}^{-6}\text{m}\\ \\ =(7.4584\times {10}^{-6}\text{m})\times \left(1\times {10}^9\text{nm/m}\right)\\ \\ =7458.4\text{nm}\end{array}$

The wavelength of lowest energy line in Pfund series of the excited $\text{H}$ atom is $7458.4nm$

Conclusion

The wavelength and frequency of the least energetic in Pfund series of the excited $\text{H}$ atom is $7471.6\text{nm}$ and $4.012\text{×}{\text{10}}^{13}{\text{s}}^{-1}$ respectively.