Chapter 7, Problem 7.60QP

(a)

Interpretation Introduction

Interpretation: The given UV line associated with a transition between different excited states or between an excited state and the ground state is to be stated. The value of $n_1$ for the transition and the energy of the longest wavelength photon are to be calculated.

Concept introduction: When an atom or a molecule makes electronic transition from a high energy state to a lower energy state, the spectrum of different frequencies of electromagnetic radiations is obtained known as emission spectrum.

To determine: If the given UV line associated with a transition between different excited states or between an excited state and the ground state.

Answer to Problem 7.60QP

Solution

An atomic emission line with a wavelength of $92.3\text{nm}$ is associated with a transition between an excited state and the ground state.

Explanation of Solution

Explanation

Given

The wavelength is $92.3\text{nm}$.

The wavelength is related to the energy level in the hydrogen atom which is given by the Rydberg equation,

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Where,

• $\lambda$ is the wavelength.
• $R$ is the Rydberg constant $\left(1.09\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\right)$.
• $n_1$ is the lower energy state.
• $n_2$ is the higher energy state.

Substitute the value of $R$ and $\lambda$ in the above equation.

$\frac{1}{\lambda }=1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

On rearranging the above equation,

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{\lambda \times 1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}}\\ \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{92.3\text{nm}\times 1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}}\\ \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{1.012}\\ \frac{1}{n_1^2}-\frac{1}{n_2^2}=0.9881\end{array}$

Simplify the above equation,

$\frac{1}{n_1^2}-\frac{1}{n_2^2}\approx 1$

This is only possible when the values of $n_1$ and $n_2$ are 1 and $\infty$ respectively. Therefore, an atomic emission line with a wavelength of $92.3\text{nm}$ is associated with a transition between an excited state and the ground state.

(b)

Interpretation Introduction

To determine: The value of $n_1$ for the transition.

Answer to Problem 7.60QP

Solution

The value of $n_1$ in the transition is 1.

Explanation of Solution

Explanation

Hence, the value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ is one. This is possible only when the value of $n_1$ is one. Therefore,

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{1^2}-\frac{1}{{\infty }^2}\\ =1-0\\ =1\end{array}$

Hence, the value of $n_1$ in the transition is 1.

(c)

Interpretation Introduction

To determine: The energy of the longest wavelength photon.

Answer to Problem 7.60QP

Solution

The energy of the longest wavelength of photon is $\underset{\_}{1.63\times 10^{-18}J}$.

Explanation of Solution

Explanation

The value of $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ must be low for the longest wavelength photon that a ground state hydrogen atom absorb because wavelength is inversely proportional to $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$.

Now, here $n_1=1$ is the ground state. For the lowest value of $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, the value of $n_2$ must be 2.

The wavelength from $n=2\text{to}n=1$ transition is calculated by using the Rydberg equation,

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Where,

• $\lambda$ is the wavelength of photons.
• $R$ is the Rydberg constant $\left(1.09\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\right)$.
• $n_1$ is the lower energy state.
• $n_2$ is the higher energy state.

The values of $n_1$ and $n_2$ are 1 and 2 respectively.

Substitute the value of $R$, $n_1$ and $n_2$ in the above equation to calculate the wavelength.

$\begin{array}{c}\frac{1}{\lambda }=1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)\\ \frac{1}{\lambda }=1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\left(\frac{1}{1}-\frac{1}{4}\right)\\ \frac{1}{\lambda }=1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\left(\frac{4-1}{1\times 4}\right)\\ \frac{1}{\lambda }=1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\times \frac{3}{4}\end{array}$

Simplify the above equation,

$\begin{array}{c}\frac{1}{\lambda }=1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\times 0.75\\ \lambda =\frac{1}{1.097\times {10}^{-2}{\left(\text{nm}\right)}^{-1}\times 0.75}\\ =\frac{1}{8.2275\times {10}^{-3}{\left(\text{nm}\right)}^{-1}}\\ =121.5\text{nm}\end{array}$

Hence, the wavelength of the photon is $121.5\text{nm}$.

The conversion of nm to m is done as,

$1\text{nm}={10}^{-9}\text{m}$

Therefore the conversion of $121.5\text{nm}$ to m is done as,

$121.5\text{nm}=121.5\times {10}^{-9}\text{m}$

The energy of the longest wavelength photon is calculated by using the formula,

$E=\frac{hc}{\lambda }$

Where,

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)$.
• $c$ is the speed of light $\left(3\times {10}^8\text{m/s}\right)$.
• $\lambda$ is the wavelength.

Substitute the values of $h$, $c$ and $\lambda$ in the above formula to calculate the energy.

$\begin{array}{c}E=\frac{\left(6.626\times {10}^{-34}\text{J}\text{.s}\right)\left(3\times {10}^8\text{m/s}\right)}{121.5\times {10}^{-9}\text{m}}\\ =\frac{1.987\times {10}^{-25}\text{J}\text{.m}}{121.5\times {10}^{-9}\text{m}}\\ =\underset{\_}{1.636\times 10^{-18}J}\end{array}$

Hence, the energy of the longest wavelength of photon is $\underset{\_}{1.63\times 10^{-18}J}$.

Conclusion

1. a. An atomic emission line with a wavelength of $92.3\text{nm}$ is associated with a transition between an excited state and the ground state.
2. b. The value of $n_1$ in the transition is 1.
3. c. The energy of the longest wavelength of photon is $\underset{\_}{1.63\times 10^{-18}J}$.