Question

A hot bar of iron glows a dull red. Using our simple ball-spring model of a solid (Figure 8.23), answer the following questions,explaining in detail the processes involved. You will need to make some rough estimates of atomic properties based on prior work. (a) What is the approximate energy of the lowest-energy spectral emission line? Give a numerical value. (b) What is the approximate energy of the highest-energy spectral emission line? Give a numerical value. (c) What is the quantum number of the highest-energy occupied state? (d) Predict the energies of two other lines in the emission spectrum of the glowing iron bar. (Note: Our simple model is too simple-the actual spectrum is more complicated. However, this simple analysis gets at some important aspects of the phenomenon.)

Short answer

(a) The approximate energy of the lowest-energy spectral emission line is$2.5\times {10}^{-20}\text{\hspace{0.17em}J}$ .

(b) The approximate energy of the highest-energy spectral emission line is $2.88\times {10}^{-19}\text{\hspace{0.17em}J}$.

(c) The quantum number of the highest energy occupied state is about $12$.

(d) The energies of two other lines in the emission spectrum are$5\times {10}^{-20}\text{\hspace{0.17em}J}$ and$7.5\times {10}^{-20}\text{\hspace{0.17em}J}$ respectively.

Step-by-step solution

Significance of the energy

The energy is referred to as a qualitative property which is transferred from one object to another object. It can also not be destroyed nor created.

(a) Determination of the approximate energy of the lowest energy line

The lowest energy emission spectrum line mainly represents the jump from one to another vibrational energy. The energy required for melting the iron is the energy of the lowest energy emission spectrum line.

The equation of the lowest energy emission spectrum line is expressed as:

$\mathrm{E}=\mathrm{kT}$

Here,$\mathrm{E}$ is the lowest energy emission spectrum line,$\mathrm{k}$ is the Boltzmann constant and$\mathrm{T}$ is the iron’s melting point.

Substitute $1.38\times {10}^{-23}\text{J/K}$for$\mathrm{k}$ and$1811\text{\hspace{0.17em}K}$ for$\mathrm{T}$ in the above equation.

$\begin{array}{c}\mathrm{E}=(1.38\times {10}^{-23}\text{J/K})\left(1811\text{\hspace{0.17em}K}\right)\\ =2.5\times {10}^{-20}\text{\hspace{0.17em}J}\end{array}$

Thus, the approximate energy of the lowest-energy spectral emission line is $2.5\times {10}^{-20}\text{\hspace{0.17em}J}$.

(b) Determination of the approximate energy of the highest energy line 

The red colour line from the diagram given in the question is the highest energy emission spectral line.

The equation of the energy of the highest emission spectral line is expressed as:

${\mathrm{E}}_1=\mathrm{hf}$

Here,${\mathrm{E}}_1$ is the energy of the highest emission spectral line,$\mathrm{h}$ is the Planck’s constant and$f$ is the red light’s frequency.

Substitute$6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for$h$ and $435\times {10}^{-12}{\text{\hspace{0.17em}s}}^{\text{-1}}$for$f$ in the above equation.

$\begin{array}{c}{\mathrm{E}}_1=(6.626\times {10}^{-34}\text{J}\cdot \text{s})(435\times {10}^{-12}{\text{\hspace{0.17em}s}}^{\text{-1}})\\ =2.88\times {10}^{-19}\text{\hspace{0.17em}J}\end{array}$

Thus, the approximate energy of the highest-energy spectral emission line is $2.88\times {10}^{-19}\text{\hspace{0.17em}J}$.

(c) Determination of the quantum number 

The equation of the quantum number is expressed as:

$\mathrm{N}=\frac{{\mathrm{E}}_1}{\mathrm{E}}$

Here,$\mathrm{N}$ is the quantum number.

Substitute the values in the above equation.

$\begin{array}{c}\mathrm{N}=\frac{2.88\times {10}^{-19}\text{\hspace{0.17em}J}}{2.5\times {10}^{-20}\text{\hspace{0.17em}J}}\\ =11.52\\ \approx 12\end{array}$

Thus, the quantum number of the highest energy occupied state is about $12$.

(d) Determination of the prediction of energies

The equation of the energy of the first line in the emission spectrum is expressed as:

$\begin{array}{c}{\mathrm{U}}_1=\mathrm{E}+\mathrm{E}\\ =2\mathrm{E}\end{array}$

Here,${\mathrm{U}}_1$ is the energy of the first line in the emission spectrum.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{U}}_1=2\times 2.5\times {10}^{-20}\text{\hspace{0.17em}J}\\ =5\times {10}^{-20}\text{\hspace{0.17em}J}\end{array}$

The equation of the energy of the second line in the emission spectrum is expressed as:

$\begin{array}{c}{\mathrm{U}}_2=\mathrm{E}+2\mathrm{E}\\ =3\mathrm{E}\end{array}$

Here,${\mathrm{U}}_2$ is the energy of the second line in the emission spectrum.

Substitute the values in the above equation.

$\begin{array}{c}{\mathrm{U}}_2=3\times 2.5\times {10}^{-20}\text{\hspace{0.17em}J}\\ =7.5\times {10}^{-20}\text{\hspace{0.17em}J}\end{array}$

Here, these calculations are accurate and also wild.

Thus, the energies of two other lines in the emission spectrum are$5\times {10}^{-20}\text{\hspace{0.17em}J}$ and$7.5\times {10}^{-20}\text{\hspace{0.17em}J}$ respectively.