Chapter 11, Problem 11A.1AE

(i)

Interpretation Introduction

Interpretation:

The ratio $A/B$ for transitions with $70.8\text{pm}$ X-rays has to be calculated.

Concept introduction:

Emission is defined as the release of energy or gas into the environment.  During emission, the transfer of energy from one radioactive substance to the other takes place.  The electron gets excited from ground state to the excited state when small amount of energy is transferred and emits electromagnetic energy when it comes back.  The formula to calculate the ratio $A/B$ for transition is given below.

    $\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}$

Answer to Problem 11A.1AE

The ratio of $A/B$ for transitions with $70.8\text{pm}$ X-rays is $\underset{\_}{0.0468J\cdot s{m}^{-3}}$.

Explanation of Solution

The formula to calculate the ratio $A/B$ for transition is given below.

    $\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}$        (1)

Where,

• $A$ is the Einstein’s coefficient of spontaneous emission.
• $B$ is the Einstein’s coefficient of stimulated emission.
• $h$ is the Planck’s constant
• $v$ is the frequency of the given radiation.
• $c$ is the speed of light.

The formula to calculate frequency is given by the expression as shown below.

    $v=\frac{c}{\lambda }$        (2)

Where,

• $c$ is the speed of light.
• $\lambda$ is the wavelength.

Substitute the value of $v$ in the equation (1).

    $\begin{array}{l}\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}\\ \frac{A}{B}=\frac{8\text{π}h}{{\lambda }^3}\end{array}$        (3)

The given value of wavelength is $70.8\text{pm}$.

The conversion of $\text{pm}$ into $\text{m}$ is done as shown below.

    $1\text{pm}={10}^{-12}\text{m}$

Therefore, the conversion of $\text{70}\text{.8}\text{pm}$ into $\text{m}$ is as follows.

    $\begin{array}{c}1\text{pm}={10}^{-12}\text{m}\\ 70.8\text{pm}=70.8\times {10}^{-12}\text{m}\end{array}$

Substitute the value of $\lambda$ in the equation (3).

    $\begin{array}{c}\frac{A}{B}=\frac{8\text{π}h}{{\lambda }^3}\\ =\frac{8\times 3.14\times \left(6.625\times {10}^{-34}\text{Js}\right)}{{\left(70.8\times {10}^{-12}\text{m}\right)}^3}\\ =\underset{\_}{0.0468J\cdot s{m}^{-3}}\end{array}$

Therefore, the ratio of $A/B$ for transitions with $70.8\text{pm}$ X-rays is $\underset{\_}{0.0468J\cdot s{m}^{-3}}$.

(ii)

Interpretation Introduction

Interpretation:

The ratio $A/B$ for transitions with $50\text{0}\text{nm}$ visible light has to be calculated.

Concept introduction:

Emission is defined as the release of energy or gas into the environment.  During emission, the transfer of energy from one radioactive substance to the other takes place.  The electron gets excited from ground state to the excited state when small amount of energy is transferred and emits electromagnetic energy when it comes back.  The formula to calculate the ratio $A/B$ for transition is given below.

    $\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}$

Answer to Problem 11A.1AE

The ratio $A/B$ for transitions with $50\text{0}\text{nm}$ visible light is $\underset{\_}{1.33\times 10^{-13}J\cdot s{m}^{-3}}$.

Explanation of Solution

The formula to calculate the ratio $A/B$ for transition is given below.

    $\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}$        (1)

Where,

• $A$ is the Einstein’s coefficient of spontaneous emission.
• $B$ is the Einstein’s coefficient of stimulated emission.
• $h$ is the Planck’s constant
• $v$ is the frequency of the given radiation.
• $c$ is the speed of light.

The formula to calculate frequency is given by the expression as shown below.

    $v=\frac{c}{\lambda }$        (2)

Where,

• $c$ is the speed of light.
• $\lambda$ is the wavelength.

Substitute the value of $v$ in the equation (1).

    $\begin{array}{l}\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}\\ \frac{A}{B}=\frac{8\text{π}h}{{\lambda }^3}\end{array}$        (3)

The given value of wavelength is $500\text{nm}$.

The conversion of $\text{nm}$ into $\text{m}$ is done as shown below.

    $1\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $\text{500}\text{nm}$ into $\text{m}$ is as follows.

    $\begin{array}{c}1\text{nm}={10}^{-9}\text{m}\\ 500\text{nm}=500\times {10}^{-9}\text{m}\end{array}$

Substitute the value of $\lambda$ in the equation (3).

    $\begin{array}{c}\frac{A}{B}=\frac{8\text{π}h}{{\lambda }^3}\\ =\frac{8\times 3.14\times \left(6.625\times {10}^{-34}\text{Js}\right)}{{\left(500\times {10}^{-9}\text{m}\right)}^3}\\ =\underset{\_}{1.33\times 10^{-13}J\cdot s{m}^{-3}}\end{array}$

Therefore, the ratio $A/B$ for transitions with $50\text{0}\text{nm}$ visible light is $\underset{\_}{1.33\times 10^{-13}J\cdot s{m}^{-3}}$.

(iii)

Interpretation Introduction

Interpretation:

The ratio $A/B$ for transitions with $3000{\text{cm}}^{-1}$ infrared radiation has to be calculated.

Concept introduction:

Emission is defined as the release of energy or gas into the environment.  During emission, the transfer of energy from one radioactive substance to the other takes place.  The electron gets excited from ground state to the excited state when small amount of energy is transferred and emits electromagnetic energy when it comes back.  The formula to calculate the ratio $A/B$ for transition is given below.

    $\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}$

Answer to Problem 11A.1AE

The ratio $A/B$ for transitions with $3000{\text{cm}}^{-1}$ infrared radiation is $\underset{\_}{4.5\times 10^{-28}J\cdot s{m}^{-3}}$.

Explanation of Solution

The formula to calculate the ratio $A/B$ for transition is given below.

    $\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}$        (1)

Where,

• $A$ is the Einstein’s coefficient of spontaneous emission.
• $B$ is the Einstein’s coefficient of stimulated emission.
• $h$ is the Planck’s constant
• $v$ is the frequency of the given radiation.
• $c$ is the speed of light.

The formula to calculate frequency is given by the expression as shown below.

    $v=\frac{c}{\lambda }$        (2)

Where,

• $c$ is the speed of light.
• $\lambda$ is the wavelength.

Substitute the value of $v$ in the equation (1).

    $\begin{array}{l}\frac{A}{B}=\frac{8\text{π}h{v}^3}{c^3}\\ \frac{A}{B}=\frac{8\text{π}h}{{\lambda }^3}\end{array}$        (3)

The formula to calculate wave number is given below.

    $\overline{v}=\frac{1}{\lambda }$

The above equation can be written as shown below.

    $\lambda =\frac{1}{\overline{v}}$        (4)

Substitute the value of $\lambda$ in the equation (4).

    $\frac{A}{B}=8\text{π}h{\overline{v}}^3$        (5)

The given value of wave number is $3000{\text{cm}}^{-1}$.

The conversion of $\text{cm}$ into $\text{m}$ is done as shown below.

    $1\text{cm}={10}^{-2}\text{m}$

Therefore, the conversion of $3000{\text{cm}}^{-1}$ into $\text{m}$ is done as shown below.

    $\begin{array}{c}1\text{cm}={10}^{-2}\text{m}\\ 3000\text{cm}=3000\times {10}^{-2}\text{m}\end{array}$

Substitute the value of wave number in the equation (5).

    $\begin{array}{c}\frac{A}{B}=8\text{π}h{\overline{v}}^3\\ =8\times 3.14\times \left(6.625\times {10}^{-34}\text{J}\cdot \text{s}\right)\times {\left(3000\times {10}^{-2}\text{m}\right)}^3\\ =\text{4493}{\text{.3×10}}^{-28}\text{J}\cdot \text{s}{\text{m}}^{-\text{3}}\\ \simeq \underset{\_}{4.5\times 10^{-28}J\cdot s{m}^{-3}}\end{array}$

Therefore, the ratio $A/B$ for transitions with $3000{\text{cm}}^{-1}$ infrared radiation is $\underset{\_}{4.5\times 10^{-28}J\cdot s{m}^{-3}}$.