91Ó°ÊÓ

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.

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 $1.38×{10}^{−23}\text{J/K}$for$\mathrm{k}$ and$1811\text{‿é}$ for$\mathrm{T}$ in the above equation.

$\begin{array}{c}\mathrm{E}=(1.38×{10}^{−23}\text{J/K})\left(1811\text{‿é}\right)\\ =2.5×{10}^{−20}\text{ J}\end{array}$

Thus, the approximate energy of the lowest-energy spectral emission line is $2.5×{10}^{−20}\text{ J}$.

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$6.626×{10}^{−34}\text{​J}â‹\dots \text{s}$ for$h$ and $435×{10}^{−12}{\text{ s}}^{\text{-1}}$for$f$ in the above equation.

$\begin{array}{c}{\mathrm{E}}_1=(6.626×{10}^{−34}\text{​J}â‹\dots \text{s})(435×{10}^{−12}{\text{ s}}^{\text{-1}})\\ =2.88×{10}^{−19}\text{ J}\end{array}$

Thus, the approximate energy of the highest-energy spectral emission line is $2.88×{10}^{−19}\text{ J}$.

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.

$\begin{array}{c}\mathrm{N}=\frac{2.88×{10}^{−19}\text{ J}}{2.5×{10}^{−20}\text{ J}}\\ =11.52\\ ≈12\end{array}$

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

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.

$\begin{array}{c}{\mathrm{U}}_1=2×2.5×{10}^{−20}\text{ J}\\ =5×{10}^{−20}\text{ J}\end{array}$

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.

$\begin{array}{c}{\mathrm{U}}_2=3×2.5×{10}^{−20}\text{ J}\\ =7.5×{10}^{−20}\text{ J}\end{array}$

Here, these calculations are accurate and also wild.

Thus, the energies of two other lines in the emission spectrum are$5×{10}^{−20}\text{ J}$ and$7.5×{10}^{−20}\text{ J}$ respectively.