Chapter 20, Problem 20.6P

(a)

Interpretation Introduction

Interpretation:

Energy difference between the ground and excited state of $\text{Ca}$ has to be calculated.

Concept Introduction:

Planck’s hypothesis states energy difference associated with photon is as follows:

  $\Delta E=\frac{hc}{\lambda }$

Here,

$\lambda$ denotes wavelength.

$h$ denotes Planck’s constant

$c$ denotes speed of light.

$\Delta E$ denotes energy difference.

Boltzmann’s distribution law for relative population is as follows:

  $\frac{N*}{N_0}=\left(\frac{g*}{g}\right)e^{-\Delta E/kT}$

Here,

$N*$ denotes population of excited state.

$N_0$ denotes population in ground state.

$g*$ denotes degeneracy of excited state.

$g$ denotes degeneracy of ground state.

$k$ denotes Boltzmann’s distribution.

$T$ denotes temperature.

$\Delta E$ denotes energy difference.

Explanation of Solution

Conversion factor to convert $\text{nm}$ to $\text{m}$ is as follows:

  $1\text{ nm}={10}^{-9}\text{ m}$

Thus, $\text{422}\text{.7 nm}$ is converted to $\text{m}$ as follows:

  $\begin{array}{c}\text{Wavelength}=\left(\text{422}\text{.7 nm}\right)\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)\\ =4.227\times {10}^{-7}\text{ m}\end{array}$

Planck’s hypothesis states energy difference associated with photon is as follows:

  $\Delta E=\frac{hc}{\lambda }$        (1)

Substitute $4.227\times {10}^{-7}\text{ m}$ for $\lambda$ , $6.626\times {10}^{-34}\text{ J s}$ for $h$ and $3\times {10}^8\text{ m/s}$ for $c$ in equation (3).

  $\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{ J s}\right)\left(3\times {10}^8\text{ m/s}\right)}{\left(4.227\times {10}^{-7}\text{ m}\right)}\\ =4.7\times {10}^{-19}\text{ J/photon}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Ratio of $N*\text{/}{N}_0$ at $2500\text{ K}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Boltzmann’s distribution law for relative population is as follows:

  $\frac{N*}{N_0}=\left(\frac{g*}{g}\right)e^{-\Delta E/kT}$        (2)

Substitute 3 for $g*\text{/}g$, $4.7\times {10}^{-19}\text{ J}$ for $\Delta E$ , $\text{1}\text{.381}\times {\text{10}}^{-23}\text{ J/K}$ for $\Delta E$ and $2500\text{ K}$ for $T$ in equation (2).

  $\begin{array}{c}\frac{N*}{N_0}=\left(\frac{3}{1}\right)e^{-\left(4.7\times {10}^{-19}\text{ J}\right)/\left(1.381\times {10}^{-23}\right)\left(2500\text{ K}\right)}\\ =3.672\times {10}^{-6}\\ \approx 3.67\times {10}^{-6}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Percent fraction change when temperature changes to $2515\text{ K}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Substitute 3 for $g*\text{/}g$, $4.7\times {10}^{-19}\text{ J}$ for $\Delta E$ , $\text{1}\text{.381}\times {\text{10}}^{-23}\text{ J/K}$ for $\Delta E$ and $2515\text{ K}$ for $T$ in equation (2).

  $\begin{array}{c}\frac{N*}{N_0}=\left(\frac{3}{1}\right)e^{-\left(4.7\times {10}^{-19}\text{ J}\right)/\left(1.381\times {10}^{-23}\right)\left(2515\text{ K}\right)}\\ =3.982\times {10}^{-6}\end{array}$

Thus percent change when temperature changes to $2515\text{ K}$ is calculated as follows:

  $\begin{array}{c}\text{Percent change}=\left(\frac{3.982\times {10}^{-6}-3.672\times {10}^{-6}}{3.672\times {10}^{-6}}\right)\left(100\text{ \%}\right)\\ =8.4\text{ \%}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Ratio of $N*\text{/}{N}_0$ at $\text{6000 K}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Substitute 3 for $g*\text{/}g$, $4.7\times {10}^{-19}\text{ J}$ for $\Delta E$ , $\text{1}\text{.381}\times {\text{10}}^{-23}\text{ J/K}$ for $\Delta E$ and $\text{6000 K}$ for $T$ in equation (2).

  $\begin{array}{c}\frac{N*}{N_0}=\left(\frac{3}{1}\right)e^{-\left(4.7\times {10}^{-19}\text{ J}\right)/\left(1.381\times {10}^{-23}\right)\left(\text{6000 K}\right)}\\ =0.0103\end{array}$

Hence, ratio of $N*\text{/}{N}_0$ is found as 0.0103.