Chapter 27, Problem 48P

To determine

(a) To Determine:

Is the transition state an absorption or emission and which state energy is higher initial or final when initial state $(n)$ is 1 and final state $(n^{\prime })$ is 3.

Answer to Problem 48P

Solution:

Final state is higher, so the transition of energy is absorption. The energy of the photon is $12.1\text{eV}$

Explanation of Solution

Given:

Initial state $(n)$ is 1

Final state $(n^{\prime })$ is 3.

Formula Used:

$hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left(\frac{1}{n^2}\right)$

Calculation:

The energy of a level can be determined by using the formula $hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left(\frac{1}{n^2}\right)$

In this question, initial state $(n)$ is 1 and final state $(n^{\prime })$ is 3. As the value of final state is higher than the value of the initial state so the transition of the energy is absorption.

The energy of the photon can be calculated as

$hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left[\left(\frac{1}{3^2}\right)-\left(\frac{1}{1^2}\right)\right]=12.1\text{eV}$

So, the energy of the photon in this transition is $12.1\text{eV}$

To determine

(b) To Determine:

Is the transition state an absorption or emission and which state energy is higher initial or final when initial state $(n)$ is 6 and final state $(n^{\prime })$ is 2.

Answer to Problem 48P

Solution:

Final state is lower than the initial state. So, the transition of energy is emission. The energy of the photon is $3.0\text{eV}$.

Explanation of Solution

Given:

Initial state $(n)$ is 6.

Final state $(n^{\prime })$ is 2.

Formula Used:

$hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left(\frac{1}{n^2}\right)$

Calculation:

The energy of a level can be determined by using the formula $hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left(\frac{1}{n^2}\right)$

In this question, initial state $(n)$ is 6 and final state $(n^{\prime })$ is 2. As the value of final state is lower than the value of the initial state so the transition of the energy is emission.

The energy of the photon can be calculated as

$hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left[\left(\frac{1}{6^2}\right)-\left(\frac{1}{2^2}\right)\right]=3.0\text{eV}$

So, the energy of the photon in this transition is $3.0\text{eV}$

To determine

(c) To Determine:

Is the transition state an absorption or emission and which state energy is higher initial or final when initial state $(n)$ is 4 and final state $(n^{\prime })$ is 5.

Answer to Problem 48P

Solution:

Final state is higher, so the transition of energy is absorption. The energy of the photon is $0.31\text{eV}$

Explanation of Solution

Given:

Initial state $(n)$ is 4

Final state $(n^{\prime })$ is 5

Formula Used:

$hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left(\frac{1}{n^2}\right)$

Calculation:

The energy of a level can be determined by using the formula $hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left(\frac{1}{n^2}\right)$

In this question, initial state $(n)$ is 4 and final state $(n^{\prime })$ is 5. As the value of final state is higher than the value of the initial state so the transition of the energy is absorption.

The energy of the photon can be calculated as

$hf=E_{n^{\prime }}-E_n=-(13.6\text{eV})\left[\left(\frac{1}{4^2}\right)-\left(\frac{1}{5^2}\right)\right]=0.31\text{eV}$

So, the energy of the photon in this transition is $0.31\text{eV}$