Chapter 7, Problem 7.119SP

(a)

Interpretation Introduction

Interpretation:

The electronic transition corresponds to line B and C has to be identified.

Explanation of Solution

The given lines are corresponding to $\text{n}\text{=}\text{2}$.  Line A has longest wavelength or lowest energy transition.  This indicates the transition is $3\underset{}{\overset{}{\to }}2$ transition.  Line B has greater wavelength and lower energy than line C.  Therefore, line B corresponds to $4\underset{}{\overset{}{\to }}2$  and line C corresponds to $5\underset{}{\overset{}{\to }}2$ transition.

(b)

Interpretation Introduction

Interpretation:

From the wavelength of line c, the wavelength of line A and B has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this, he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \\ \text{Eis}\text{energy}\text{level}\text{.}\\ \\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Answer to Problem 7.119SP

The wavelength of line A is $41.1\text{nm}$

The wavelength of line B is $\text{30}\text{.4nm}$

Explanation of Solution

The energy of line C is calculated using its wavelength as follows,

$\begin{array}{l}\text{E}\text{=}\frac{\text{hc}}{\text{λ}}\text{=}\frac{\text{(6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s)(3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s)}}{\text{(27}\text{.1}\text{×}{\text{10}}^{\text{-9}}\text{m)}}\\ \\ \text{=}\text{7}\text{.34}\text{×}{\text{10}}^{\text{-18}}\text{J}\end{array}$

The atom in which electronic transition of $5\underset{}{\overset{}{\to }}2$ occurs is identified as

$\begin{array}{l}\text{ΔE}\text{=}\text{hν}\text{=}{\text{E}}_{\text{f}}\text{-}{\text{E}}_{\text{i}}\\ \\ \text{ΔE}\text{=}{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}\left(\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\right)\\ \text{-7}\text{.34}\text{×}{\text{10}}^{\text{-18}}\text{J}\text{=}{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}\left(\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\right)\\ \\ \text{-7}\text{.34}\text{×}{\text{10}}^{\text{-18}}\text{J}\text{=}\text{(2}\text{.18}\text{×}{\text{10}}^{\text{-18}}{\text{J)Z}}^{\text{2}}\left(\frac{\text{1}}{{\text{5}}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)\\ \\ \text{-7}\text{.34}\text{×}{\text{10}}^{\text{-18}}\text{J}\text{=}{\text{Z}}^{\text{2}}\left(\frac{\text{2}\text{.18}\text{×}{\text{10}}^{\text{-18}}\text{J}}{{\text{5}}^{\text{2}}}\text{-}\frac{\text{2}\text{.18}\text{×}{\text{10}}^{\text{-18}}\text{J}}{{\text{2}}^{\text{2}}}\right)\\ \text{-7}\text{.34}\text{×}{\text{10}}^{\text{-18}}\text{J}\text{=}{\text{Z}}^{\text{2}}\left(\frac{8.72\text{×}{\text{10}}^{\text{-18}}\text{J}\text{-}5.45\text{×}{\text{10}}^{\text{-17}}\text{J}}{100}\right)\\ \text{-7}\text{.34}\text{×}{\text{10}}^{\text{-18}}\text{J}\text{=}\text{(-4}\text{.58}\text{×}{\text{10}}^{\text{-19}}{\text{)Z}}^{\text{2}}\\ \\ {\text{Z}}^{\text{2}}\text{= 16}\text{.0}\\ \\ \text{Z}\text{=}\text{4}\end{array}$

The energy change and wavelength for transitions of $3\underset{}{\overset{}{\to }}2$ and $4\underset{}{\overset{}{\to }}2$ are calculated as

$\begin{array}{l}\text{ΔE}\text{=}{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}\left(\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\right)\\ \\ \text{=}\text{(2}\text{.18}\text{×}{\text{10}}^{\text{-18}}\text{J)}{\text{(4)}}^{\text{2}}\left(\frac{\text{1}}{3^{\text{2}}}\text{-}\frac{\text{1}}{2^{\text{2}}}\right)\\ \\ \text{=}\text{(3}\text{.488}\text{×}{\text{10}}^{\text{-17}}\text{J)}\left(\frac{\text{1}}{3^{\text{2}}}\text{-}\frac{\text{1}}{2^{\text{2}}}\right)\\ \\ \text{ =}\left(\frac{\text{3}\text{.488}\text{×}{\text{10}}^{\text{-17}}\text{J}}{3^{\text{2}}}\text{-}\frac{\text{3}\text{.488}\text{×}{\text{10}}^{\text{-17}}\text{J}}{2^{\text{2}}}\right)\\ \\ \text{ =}\left(\frac{\text{1}\text{.395}\text{×}{\text{10}}^{\text{-16}}\text{J- 3}\text{.139}\text{×}{\text{10}}^{\text{-16}}\text{J }}{36}\right)\\ \\ \text{=}\text{-4}\text{.84}\text{×}{\text{10}}^{\text{-18}}\text{J}\end{array}$

Negative sign is neglected while calculating lambda

$\begin{array}{l}\text{λ}\text{=}\frac{\text{hc}}{\text{ΔE}}\text{=}\frac{\text{(6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s)(3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s)}}{\text{4}\text{.84}\text{×}{\text{10}}^{\text{-18}}\text{J}}\\ \\ \text{λ}\text{=}4.11\text{×}{\text{10}}^{\text{-8}}\text{m}\text{=}41.1\text{nm}\text{1m}\text{=}{\text{10}}^{\text{9}}\text{nm}\end{array}$

$\begin{array}{l}\text{ΔE}\text{=}{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}\left(\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\right)\\ \\ \text{=}\text{(2}\text{.18}\text{×}{\text{10}}^{\text{-18}}\text{J)}{\text{(4)}}^{\text{2}}\left(\frac{\text{1}}{4^{\text{2}}}\text{-}\frac{\text{1}}{2^{\text{2}}}\right)\\ \\ =3.488\text{×}{\text{10}}^{\text{-17}}\left(\frac{-3}{16}\right)\\ \\ \text{=}\text{-6}\text{.54}\text{×}{\text{10}}^{\text{-18}}\text{J}\end{array}$

Negative sign is neglected while calculating lambda

$\begin{array}{l}\text{λ}\text{=}\frac{\text{hc}}{\text{ΔE}}\text{=}\frac{\text{(6}\text{.63}\text{×}{\text{10}}^{\text{-34}}\text{J}\text{.s)(3}\text{.00}\text{×}{\text{10}}^{\text{8}}\text{m/s)}}{\text{6}\text{.54}\text{×}{\text{10}}^{\text{-18}}\text{J}}\\ \\ \text{λ}\text{=}\text{3}\text{.04}\text{×}{\text{10}}^{\text{-8}}\text{m}\text{=}\text{30}\text{.4nm}(\text{1m}\text{=}{\text{10}}^{\text{9}}\text{nm)}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The energy required to eject an electron from $\text{n}\text{=}\text{4}$ has to be calculated.

Concept Introduction:

Bohr developed a rule for quantization of energy that could be applicable to the electron of an atom in motion.  By using this, he derived a formula for energy levels of electron in H-atom.

$\begin{array}{l}\text{E}\text{=}\frac{{\text{-R}}_{\text{H}}}{{\text{n}}^{\text{2}}}\text{n}\text{=}\text{1,2,3,}\mathrm{......}\text{(For}\text{hydrogen}\text{atom)}\\ \\ {\text{R}}_{\text{H}}\text{is}\text{Rydberg}\text{constant}\text{(2}\text{.179}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{.}\\ \\ \text{Eis}\text{energy}\text{level}\text{.}\\ \\ \text{n}\text{is}\text{principal}\text{quantum}\text{number}\text{.}\end{array}$

Answer to Problem 7.119SP

The energy required to eject an electron from $\text{n}\text{=}\text{4}$ is $2.18\text{×}{\text{10}}^{\text{-18}}\text{J}$

Explanation of Solution

The initial state is $\text{n}\text{=}\text{4}$ and final state is infinity.  The energy change for transitions of $4\underset{}{\overset{}{\to }}\infty$ is calculated as

$\begin{array}{l}\text{ΔE}\text{=}{\text{R}}_{\text{H}}{\text{Z}}^{\text{2}}\left(\frac{\text{1}}{{\text{n}}_{\text{i}}{}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{n}}_{\text{f}}{}^{\text{2}}}\right)\\ \\ \text{=}\text{(2}\text{.18}\text{×}{\text{10}}^{\text{-18}}\text{J)}{\text{(4)}}^{\text{2}}\left(\frac{\text{1}}{4^{\text{2}}}\text{-}\frac{\text{1}}{{\infty }^{\text{2}}}\right)\\ \\ \text{=}\text{(2}\text{.18}\text{×}{\text{10}}^{\text{-18}}\text{J)}{\text{(4)}}^{\text{2}}\left(\frac{\text{1}}{16}\text{-}0\right)\\ \\ \text{=}\text{(2}\text{.18}\text{×}{\text{10}}^{\text{-18}}\text{J)}\text{(16)}\left(\frac{\text{1}}{16}\right)\\ \\ \text{=}2.18\text{×}{\text{10}}^{\text{-18}}\text{J}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The physical significance of continuum has to be explained.

Explanation of Solution

The energy levels are closely packed when the n values become larger and it leads to continuum of lines.  Electrons have been removed from atom when the continuum starts.  Therefore, there will be no quantized energy levels associated with electron.