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Q69P

Expert-verifiedFound in: Page 1218

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).

**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.

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

So, the change in energy is given by,

$\u2206E={E}_{{n}_{f}}-{E}_{{n}_{i}}\phantom{\rule{0ex}{0ex}}=-\frac{13.6eV}{{{n}_{f}}^{2}}+\frac{13.6eV}{{{n}_{i}}^{2}}\phantom{\rule{0ex}{0ex}}\u2206E=-13.6eV\left(\frac{1}{{n}_{f}^{2}}-\frac{1}{{n}_{i}^{2}}\right)$

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

$\u2206E=hv................\left(1\right)$

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

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,

$\u2206E=-13.6eV\left(\frac{1}{{n}_{f}^{2}}-\frac{1}{n{\text{'}}^{2}}\right)-13.6eV\left(\frac{1}{n{\text{'}}^{2}}-\frac{1}{{n}_{i}^{2}}\right)\phantom{\rule{0ex}{0ex}}\u2206E=\left(\u2206{E}_{1}\right)+\left(\u2206{E}_{2}\right)..........\left(2\right)$

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-

$\u2206{E}_{1}=h{v}_{1}and\u2206{E}_{2}=h{v}_{2}.................\left(3\right)$

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

$\u2206E=\left(\u2206{E}_{1}\right)+\left(\u2206{E}_{1}\right)\phantom{\rule{0ex}{0ex}}hv=\left(h{v}_{1}\right)+\left(h{v}_{2}\right)\phantom{\rule{0ex}{0ex}}v={v}_{1}+{v}_{2}$

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-

$\u2206{E}_{1}=\left(-13.6eV\right)-\left(-3.4eV\right)\phantom{\rule{0ex}{0ex}}=-10.2eV\phantom{\rule{0ex}{0ex}}\u2206{E}_{2}=\left(-3.4\right)-\left(-0.51\right)\phantom{\rule{0ex}{0ex}}=-10.2eV\phantom{\rule{0ex}{0ex}}\u2206E=\left(-13.6eV\right)-\left(-0.51eV\right)\phantom{\rule{0ex}{0ex}}=-13.09eV\phantom{\rule{0ex}{0ex}}$

These values are related as-

$\u2206E=\u2206{E}_{1}+\u2206{E}_{2}\phantom{\rule{0ex}{0ex}}=-10.2eV-2.89eV\phantom{\rule{0ex}{0ex}}=13.09eV$

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.

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).

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