Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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.

$n' = 3 \to n_{f} = 2$ 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 $n_{i} = 4 \to n' = 3$ (first Paschen) followed by $n' = 3 \to n_{f} = 2$ (first Balmer).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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,

$∆ E = E_{n_{f}} - E_{n_{i}} = - \frac{13.6 e V}{{n_{f}}^{2}} + \frac{13.6 e V}{{n_{i}}^{2}} ∆ E = - 13.6 e V (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$

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

$∆ E = h v . . . . . . . . . . . . . . . . (1)$

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 $n_{i} \to n'$ to $n' \to n_{f}$

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

$∆ E = - 13.6 e V (\frac{1}{n_{f}^{2}} - \frac{1}{n'^{2}}) - 13.6 e V (\frac{1}{n'^{2}} - \frac{1}{n_{i}^{2}}) ∆ E = (∆ E_{1}) + (∆ E_{2}) . . . . . . . . . . (2)$

Here, we have two transitions taking place, so two photons having energies $h v_{1} a n d h v_{2}$ are released having frequencies role="math" localid="1661856051342" $v_{1} a n d v_{2}$ respectively. Also, we have-

$∆ E_{1} = h v_{1} a n d ∆ E_{2} = h v_{2} . . . . . . . . . . . . . . . . . (3)$

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

$∆ E = (∆ E_{1}) + (∆ E_{1}) h v = (h v_{1}) + (h v_{2}) v = v_{1} + v_{2}$

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

$∆ E_{1} = (- 13.6 e V) - (- 3.4 e V) = - 10.2 e V ∆ E_{2} = (- 3.4) - (- 0.51) = - 10.2 e V ∆ E = (- 13.6 e V) - (- 0.51 e V) = - 13.09 e V$

These values are related as-

$∆ E = ∆ E_{1} + ∆ E_{2} = - 10.2 e V - 2.89 e V = 13.09 e V$

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 $n_{i} = 4 \to n' = 3$ (first Paschen line) followed by $n' = 3 \to n_{f} = 2$ (first Balmer line).

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Short Answer

Step by Step Solution

Step 1: energy level diagram for hydrogen

Step 2: Explanation

Step 3: Finding another example

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