Chapter 7, Problem 7.50QP

(a)

Interpretation Introduction

Interpretation: The given transitions in the hydrogen atom in order of increasing frequency of the electromagnetic radiation are to be ranked.

Concept introduction: Frequency of light is defined as the number of waves passes in space during a given interval of time of one second. The energy of a photon is defined as the energy of a single photon. The frequency of photon is inversely proportional to the wavelength of the electromagnetic wave.

To determine: The frequency of the given transition.

Answer to Problem 7.50QP

Solution

The value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is $\underset{\_}{0.0347}$.

Explanation of Solution

Explanation

Given

The transition is from $n=4\text{to}n=6$.

The relation between the wavelength of an electronic transition and the value of orbit in a hydrogen atom is given by Rydberg equation,

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

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 frequency of photons is given by the formula,

$\nu =\frac{c}{\lambda }$ (2)

Where,

• $\nu$ is the frequency of electromagnetic radiation.
• $c$ is the speed of light.
• $\lambda$ is the wavelength.

Therefore, from equation (1) and (2) it is clear that the frequency of photons is inversely proportional to the wavelength of photons.

$\nu \alpha \frac{1}{\lambda }$

Hence,

$\nu \alpha \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Therefore, the value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is calculated as,

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{{\left(4\right)}^2}-\frac{1}{{\left(6\right)}^2}\\ =\frac{1}{16}-\frac{1}{36}\\ =\frac{36-16}{16\times 36}\\ =\frac{20}{576}\end{array}$

Simplify the above equation.

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{20}{576}\\ =\underset{\_}{0.0347}\end{array}$

Hence, the value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is $\underset{\_}{0.0347}$.

(b)

Interpretation Introduction

To determine: The frequency of the given transition.

Answer to Problem 7.50QP

Solution

The value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is $\underset{\_}{0.0121}$.

Explanation of Solution

Explanation

Given

The transition is from $n=6\text{to}n=8$.

The value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is calculated as,

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{{\left(6\right)}^2}-\frac{1}{{\left(8\right)}^2}\\ =\frac{1}{36}-\frac{1}{64}\\ =\frac{64-36}{36\times 64}\\ =\frac{28}{2304}\end{array}$

Simplify the above equation.

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{28}{2304}\\ =\underset{\_}{1.0121}\end{array}$

Hence, the value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is $\underset{\_}{0.0121}$.

(c)

Interpretation Introduction

To determine: The frequency of the given transition.

Answer to Problem 7.50QP

Solution

The value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is $\underset{\_}{0.0040}$.

Explanation of Solution

Explanation

Given

The transition is from $n=9\text{to}n=11$.

The value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is calculated as,

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{{\left(9\right)}^2}-\frac{1}{{\left(11\right)}^2}\\ =\frac{1}{81}-\frac{1}{121}\\ =\frac{121-81}{81\times 121}\\ =\frac{40}{9801}\end{array}$

Simplify the above equation.

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{40}{9801}\\ =\underset{\_}{0.0040}\end{array}$

Hence, the value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is $\underset{\_}{0.0040}$.

(d)

Interpretation Introduction

To determine: The frequency of the given transition.

Answer to Problem 7.50QP

Solution

The value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is $\underset{\_}{0.0023}$.

Explanation of Solution

Explanation

Given

The transition is from $n=11\text{to}n=13$.

The value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is calculated as,

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{{\left(11\right)}^2}-\frac{1}{{\left(13\right)}^2}\\ =\frac{1}{121}-\frac{1}{169}\\ =\frac{169-121}{121\times 169}\\ =\frac{48}{20449}\end{array}$

Simplify the above equation.

$\begin{array}{c}\frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{48}{20449}\\ =\underset{\_}{0.0023}\end{array}$

Hence, the value of $\frac{1}{n_1^2}-\frac{1}{n_2^2}$ for the given transition is $\underset{\_}{0.0023}$.

Conclusion

The transitions in the hydrogen atom in order of increasing frequency of the electromagnetic radiation are,

$n=11\text{to}n=13<n=9\text{to}n=11<n=6\text{to}n=8<n=4\text{to}n=6$.