Chapter 12, Problem 161CP

(a)

Interpretation Introduction

Interpretation:Quantum numbers for first three energy levels should be determined.

Concept introduction:Energy, size, shape, and orientation of atomic orbital are determined with help of some numbers. These numbers are called quantum numbers and are obtained from solution of Schrodinger equation of hydrogen atom by application of boundary conditions.Principal quantum number is represented by $n$ . This quantum number is related to size and energy of different atomic orbitals. With increase in value of $n$ , size of orbital becomes larger and electron is present farther from atomic nucleus for longer time.

Explanation of Solution

Value of $n$ gives the information about the energy levels in two dimension box. Value of $n$ can never be zero because it is not possible that there is no energy level. Hence, quantum numbers for first three allowed energy levels are as follows:

  $\begin{array}{c}\text{Ground state}:n_x=1,n_y=1\\ \text{First excited state}:n_x=2,n_y=1\\ \text{Second}\text{excited state}:n_x=1,n_y=2\end{array}$ ,

(b)

Interpretation Introduction

Interpretation:Wavelength of light that promotes from first excited state to second is to be calculated.

Concept introduction:Wavefunction is defined as the function of wave that gives information about the properties of wave. For a particle in one dimension box energy is calculated as follows:

  $E=\frac{n^2{h}^2}{8m{L}^2}$

Where,

• $E$ is energy of state.
• $n$ is energy level.
• $h$ isplanck’s constant.
• $m$ is mass of electron.
• $L$ is dimension of cubical box.

Answer to Problem 161CP

Value of $\lambda$ is $4.4\times {10}^{-5}\text{ m}$ .

Explanation of Solution

There are two possibilities of energy state of first excieted and second excited state that is $E_{12}$ or $E_{21}$ . Therefore, expression to calculate energy $E_{12}$ is as follows:

  $E_{12}=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)$

Where,

• $E_{12}$ is energy state for quantum number 1 and 2.
• $h$ isplanck’s constant.
• $m$ is mass of electron.
• $n_x$ and $n_y$ are energy levels in $x$ and $y$ direction respectively.
• $L_x$ and $L_y$ is dimension of box.

Value of $n_x$ is 1.

Value of $n_y$ is 2.

Value of $L_x$ is $8\text{ nm}$ .

Value of $L_y$ is $5\text{ nm}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{12}=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)\\ =\frac{h^2}{8m}\left(\frac{1}{{\left(8\text{ nm}\right)}^2}+\frac{2^2}{{\left(5\text{ nm}\right)}^2}\right){\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)}^2\\ =\frac{h^2}{8m}\left(1.76\times {10}^{17}\right)\end{array}$

Expression to calculate energy $E_{21}$ is as follows:

  $E_{21}=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)$

Where,

• $E_{21}$ is energy state for quantum number 1 and 2.
• $h$ isplanck’s constant.
• $m$ is mass of electron.
• $n_x$ and $n_y$ are energy levels in $x$ and $y$ direction respectively.
• $L_x$ and $L_y$ is dimension of box.

Value of $n_x$ is 2.

Value of $n_y$ is 1.

Value of $L_x$ is $8\text{ nm}$ .

Value of $L_y$ is $5\text{ nm}$ .

Substitute values in above equation.

  $\begin{array}{c}{E}_{21}=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)\\ =\frac{h^2}{8m}\left(\frac{2^2}{{\left(8\text{ nm}\right)}^2}+\frac{1}{{\left(5\text{ nm}\right)}^2}\right){\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)}^2\\ =\frac{h^2}{8m}\left(1.03\times {10}^{17}\right)\end{array}$

The difference in both energies results into energy difference and is calculated as follows:

  $\begin{array}{c}\Delta E=\frac{h^2}{8m}\left(1.76\times {10}^{17}\right)-\frac{h^2}{8m}\left(1.03\times {10}^{17}\right)\\ =\frac{h^2}{8m}\left(1.76\times {10}^{17}-1.03\times {10}^{17}\right)\\ =\frac{h^2}{8m}\left(0.73\times {10}^{17}\right)\end{array}$

The expression for energy difference is as follows:

  $\Delta E=\frac{h^2}{8m}\left(0.73\times {10}^{17}\right)$

Value of $h$ is $6.626\times {10}^{-34}\text{ Js}$ .

Value of $m$ is $9.1\times {10}^{-31}\text{ kg}$ .

Substitute values in above equation.

  $\begin{array}{c}\Delta E=\frac{h^2}{8m}\left(0.73\times {10}^{17}\right)\\ =\frac{{\left(6.626\times {10}^{-34}\text{ Js}\right)}^2}{8\left(9.1\times {10}^{-31}\text{ kg}\right)}\left(0.73\times {10}^{17}\right)\\ =4.4\times {10}^{-21}\text{ J}\end{array}$

The change in energy for transition is calculated is as follows:

  $\Delta E=\frac{hc}{\lambda }$

Rearrange above equation for $\lambda$ .

  $\lambda =\frac{hc}{\Delta E}$

Where,

• $\Delta E$ is energy change.
• $h$ isplanck’s constant.
• $c$ is speed of light.
• $\lambda$ is wavelength of energy.

Value of $c$ is $2.998\times {10}^8\text{ m/s}$ .

Value of $\Delta E$ is $4.4\times {10}^{-21}\text{ J}$ .

Value of $h$ is $6.626\times {10}^{-34}\text{ Js}$ .

Substitute values in above equation.

  $\begin{array}{c}\lambda =\frac{hc}{\Delta E}\\ =\frac{\left(6.626\times {10}^{-34}\text{ Js}\right)\left(2.998\times {10}^8\text{ m/s}\right)}{4.4\times {10}^{-21}\text{ J}}\left(\frac{{10}^9\text{ nm}}{1\text{ m}}\right)\\ =4.4\times {10}^{-5}\text{ m}\end{array}$

Hence, value of $\lambda$ is $4.4\times {10}^{-5}\text{ m}$ .