Chapter 1, Problem 1.2E

(a)

Interpretation Introduction

Interpretation:

Number of spectral lines appear in the spectrum as the excited atom falls back from $\text{n}\text{=}5$ to lower states has to be given.

Explanation of Solution

The highest electronic state is $\text{n}\text{=}5$ so the electronic transition from $\text{n}\text{=}5,4,3,2$ is possible to each of the corresponding lower electronic states

  $\begin{array}{l}5\to 4,5\to 3,5\to 2,5\to 1\\ 4\to 3,4\to 2,4\to 1\\ 3\to 2,3\to 1\\ 2\to 1\end{array}$

Therefore the total number of emission lines is $10$, if only five quantum levels are considered.

(b)

Interpretation Introduction

Interpretation:

Range of wavelengths emitted the electron in a hydrogen atom from $\text{n}\text{=}5$ to lower states has to be determined.

Concept Introduction:

The relationship between wavelength, frequency and speed of light is given below.

  $\lambda \times \nu =c$

Where, $\lambda$ is the wavelength, $\nu$ is the frequency of radiation and $c$ is the speed of light.

Rydberg equation:

  $\begin{array}{l}\nu =R\left\{\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right\}\\ \\ Where,n_1=1,2,\mathrm{....},n_2=n_1+1,n_1+2,\mathrm{....}\end{array}$

R is the Rydberg constant.  Its value is $3.29\times {10}^{15}Hz.$

Explanation of Solution

Here, 10 transitions are possible

  $\begin{array}{l}5\to 4,5\to 3,5\to 2,5\to 1\\ 4\to 3,4\to 2,4\to 1\\ 3\to 2,3\to 1\\ 2\to 1\end{array}$

As the energy gap between two transition states increases the wavelength of the radiation emitted decreases.

The electronic transition from the highest level to next lower electronic state has the longest wavelength.

Therefore, in a transition, the photons with longest wavelength are emitted from $\text{n}\text{=}5$ to the $\text{n}\text{=}4$ level. Also, the photons with least wavelength are emitted from $\text{n}\text{=}5$ to the $\text{n}\text{=}1$ level.

• Calculation of wavelength for the transition taking place from $\text{n}\text{=}5$ to the $\text{n}\text{=}4$ level:

  $\begin{array}{c}\nu =R\left\{\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right\}=\left(3.29\times {10}^{15}{s}^{-1}\right)\left\{\frac{1}{4^2}-\frac{1}{5^2}\right\}\\ \\ =2.7965\times {10}^{14}{s}^{-1}\\ \\ \lambda =\frac{c}{\nu }=\frac{\left(3\times {10}^{8}m/s\right)}{\left(2.7965\times {10}^{14}{s}^{-1}\right)}\\ \\ =1.07\times {10}^{-6}m=1073nm\end{array}$

The longest wavelength of light emitted is $1073nm$.

Calculation of wavelength for the transition taking place from $\text{n}\text{=}5$ to the $\text{n}\text{=}1$ level:

  $\begin{array}{c}\nu =R\left\{\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right\}\\ \\ =R\left\{\frac{1}{1^2}-\frac{1}{5^2}\right\}=\frac{24}{25}R\\ \\ \lambda =\frac{c}{\nu }=\frac{c}{\frac{24}{25}R}=\frac{25c}{24R}\\ \\ =\frac{25\times \left(3\times {10}^{8}m/s\right)}{24\times \left(3.29\times {10}^{15}{s}^{-1}\right)}\\ \\ =9.50\times {10}^{-7}m=95.0nm\end{array}$

The least wavelength of light emitted is $\text{95}\text{.0}\text{nm}$.

Therefore, range of wavelengths emitted is from $1073nm$ to $95.0nm$.