Chapter 21, Problem 21.16P

(a)

Interpretation Introduction

Interpretation:

The energy difference between the ground and excited state has to be calculated.

Answer to Problem 21.16P

The energy difference between ground and excited state is $\text{283}\text{kJ/mol}$.

Explanation of Solution

We know that,

$\begin{array}{l}\Delta E=h\nu \\ =\frac{hc}{\lambda }\\ =\frac{\left(6.626\times {10}^{-34}\text{Js}\right)\left(2.998\times {10}^8\text{m/s}\right)}{422.7\times {10}^{-9}\text{m}}\\ =4.699\times {10}^{-19}\text{J/molecule}\\ \text{=}\text{283}\text{kJ/mol}\end{array}$

(b)

Interpretation Introduction

Interpretation:

${\text{N*/N}}_{\text{0}}$ at $\text{2500}\text{K}$ has to be calculated.

Answer to Problem 21.16P

${\text{N*/N}}_{\text{0}}$ is calculated as $3.67\times {10}^{-6}$.

Explanation of Solution

Temperature is given as $\text{2500}\text{K}$.

We know that,

$\Delta E=4.699\times {10}^{-19}\text{J}$

To find ${\text{N*/N}}_{\text{0}}$:

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =\frac{g*}{g_0}{e}^{-\left(4.699\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(2500\text{K}\right)}\\ =3.67\times {10}^{-6}\end{array}$

(c)

Interpretation Introduction

Interpretation:

${\text{N*/N}}_{\text{0}}$ at $\text{2515}\text{K}$ has to be calculated and how much percentage it is increased when temperature is increased from $\text{2500}\text{K}$ to $\text{2515}\text{K}$ has to be given.

Answer to Problem 21.16P

${\text{N*/N}}_{\text{0}}$ is calculated as $3.98\times {10}^{-6}$ and there is a $8.4\%$ increase when temperature is raised from $\text{2500}\text{K}$ to $\text{2515}\text{K}$

Explanation of Solution

Temperature is given as $\text{2515}\text{K}$.

We know that,

$\Delta E=4.699\times {10}^{-19}\text{J}$

To find ${\text{N*/N}}_{\text{0}}$:

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =\frac{g*}{g_0}{e}^{-\left(4.699\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(2515\text{K}\right)}\\ =3.98\times {10}^{-6}\end{array}$

At $\text{2500}\text{K}$, ${\text{N*/N}}_{\text{0}}$ is found as $3.67\times {10}^{-6}$.

To find the percentage increase,

$\left(\frac{3.98}{3.67}\times 100\%\right)-100=8.4\%$

Therefore, there is a $8.4\%$ increase when temperature is raised by $\text{15}\text{K}$.

(d)

Interpretation Introduction

Interpretation:

${\text{N*/N}}_{\text{0}}$ at $\text{6000 K}$ has to be calculated.

Answer to Problem 21.16P

${\text{N*/N}}_{\text{0}}$ is calculated as $1.03\times {10}^{-2}$.

Explanation of Solution

Temperature is given as $\text{6000 K}$.

We know that,

$\Delta E=4.699\times {10}^{-19}\text{J}$

To find ${\text{N*/N}}_{\text{0}}$:

$\begin{array}{l}\frac{N*}{N_0}=\frac{g*}{g_0}{e}^{-\Delta E/kT}\\ =\frac{g*}{g_0}{e}^{-\left(4.699\times {10}^{-19}\text{J}\right)/\left(1.381\times {10}^{-23}\text{J/K}\right)\left(6000\text{K}\right)}\\ =1.03\times {10}^{-2}\end{array}$