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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# From the energy-level diagram for hydrogen, explain the observation that the frequency of the second Lyman-series line is the sum of the frequencies of the first Lyman-series line and the first Balmer-series line. This is an example of the empirically discovered Ritz combination principle. Use the diagram to find some other valid combinations.

$n\text{'}=3\to {n}_{f}=2$It is shown that how the frequency of the second Lyman-series line is the sum of the frequencies of the first Lyman-series line and the first Balmer-series.

Another example of such transition is the transition ${n}_{i}=4\to {n}_{f}=2$ associated with the second Balmer-series line can be thought of as ${n}_{i}=4\to n\text{'}=3$ (first Paschen) followed by $n\text{'}=3\to {n}_{f}=2$ (first Balmer).

See the step by step solution

## Step 1: energy level diagram for hydrogen

The primary quantum number, " n," has a direct impact on the energy level diagram. To calculate the energy difference between two states and describe the hydrogen spectrum, an energy diagram is required.

The energy level diagram of the hydrogen atom is displayed in the diagram below.

## Step 2: Explanation

Let the transition is from $n={n}_{i}$ to $n={n}_{f}$

So, the change in energy is given by,

$∆E={E}_{{n}_{f}}-{E}_{{n}_{i}}\phantom{\rule{0ex}{0ex}}=-\frac{13.6eV}{{{n}_{f}}^{2}}+\frac{13.6eV}{{{n}_{i}}^{2}}\phantom{\rule{0ex}{0ex}}∆E=-13.6eV\left(\frac{1}{{n}_{f}^{2}}-\frac{1}{{n}_{i}^{2}}\right)$

This decrease in energy emerges in the form of radiation, having energy hv , such that-

$∆E=hv................\left(1\right)$

Here, v is the frequency of photon released during transition and h is Planck’s constant, whose value is $6.626×{10}^{-34}J.s$ .

Now, the transition can be done in another way also.

Which is ${n}_{i}\to n\text{'}$ to $n\text{'}\to {n}_{f}$

So, by equation (1) change in energy for ${n}_{i}\to n\text{'}$ to $n\text{'}\to {n}_{f}$ is given by,

$∆E=-13.6eV\left(\frac{1}{{n}_{f}^{2}}-\frac{1}{n{\text{'}}^{2}}\right)-13.6eV\left(\frac{1}{n{\text{'}}^{2}}-\frac{1}{{n}_{i}^{2}}\right)\phantom{\rule{0ex}{0ex}}∆E=\left(∆{E}_{1}\right)+\left(∆{E}_{2}\right)..........\left(2\right)$

Here, we have two transitions taking place, so two photons having energies $h{v}_{1}andh{v}_{2}$ are released having frequencies role="math" localid="1661856051342" ${v}_{1}and{v}_{2}$ respectively. Also, we have-

$∆{E}_{1}=h{v}_{1}and∆{E}_{2}=h{v}_{2}.................\left(3\right)$

Now, by equating equation (1), (2) and (3) we get,

$∆E=\left(∆{E}_{1}\right)+\left(∆{E}_{1}\right)\phantom{\rule{0ex}{0ex}}hv=\left(h{v}_{1}\right)+\left(h{v}_{2}\right)\phantom{\rule{0ex}{0ex}}v={v}_{1}+{v}_{2}$

So, the transition ${n}_{i}=3\to {n}_{f}=1$ associated with the second Lyman-series line can be thought of as ${n}_{i}=3\to n\text{'}=2$ (first Balmer) followed by $n\text{'}=2\to {n}_{f}=1$ (first Lyman). From the energy level diagram, we have-

$∆{E}_{1}=\left(-13.6eV\right)-\left(-3.4eV\right)\phantom{\rule{0ex}{0ex}}=-10.2eV\phantom{\rule{0ex}{0ex}}∆{E}_{2}=\left(-3.4\right)-\left(-0.51\right)\phantom{\rule{0ex}{0ex}}=-10.2eV\phantom{\rule{0ex}{0ex}}∆E=\left(-13.6eV\right)-\left(-0.51eV\right)\phantom{\rule{0ex}{0ex}}=-13.09eV\phantom{\rule{0ex}{0ex}}$

These values are related as-

$∆E=∆{E}_{1}+∆{E}_{2}\phantom{\rule{0ex}{0ex}}=-10.2eV-2.89eV\phantom{\rule{0ex}{0ex}}=13.09eV$

Hence, the frequency of second Lyman-series line is the sum of the frequency of first Balmer series line and the first Lyman-series line.

## Step 3: Finding another example

From the energy level diagram, we can observe that the transition ${n}_{i}=4\to {n}_{f}=2$ associated with the second Balmer-series line can be thought of as ${n}_{i}=4\to n\text{'}=3$ (first Paschen line) followed by $n\text{'}=3\to {n}_{f}=2$ (first Balmer line).