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Chapter 6: Problem 91

Calculate the energy changes corresponding to the transitions of the hydrogen atom: (a) from \(n=3\) to \(n=4 ;\) (b) from \(n=2\) to \(n=1 ;\) and \((\mathrm{c})\) from \(n=3\) to \(n=\infty\).

Short Answer

Expert verified
The energy changes for the hydrogen atom transitions are: (a) from n=3 to n=4: \( - 0.66 \ \text{eV} \), (b) from n=2 to n=1: \( -10.2 \ \text{eV} \), and (c) from n=3 to n=∞: \( 1.51 \ \text{eV} \).

Step by step solution

01

Calculate the initial and final energy levels for each transition

Using the Rydberg formula, we can calculate the initial and final energy levels for each transition as follows: (a) \( \ E_3 = -\frac{13.6 \ \text{eV}}{3^2} = - \frac{13.6}{9} \ \text{eV} \) and \( \ E_4 = -\frac{13.6 \ \text{eV}}{4^2} = -\frac{13.6}{16} \ \text{eV} \) (b) \( \ E_2 = -\frac{13.6 \ \text{eV}}{2^2} = -\frac{13.6}{4} \ \text{eV} \) and \( \ E_1 = -\frac{13.6 \ \text{eV}}{1^2} = -13.6 \ \text{eV} \) (c) \( \ E_3 = -\frac{13.6 \ \text{eV}}{3^2} = -\frac{13.6}{9} \ \text{eV} \) and \( \ E_{\infty} = -\frac{13.6 \ \text{eV}}{\infty^2} = 0 \ \text{eV} \)
02

Calculate the energy changes for each transition

Now, we can find the energy change during the transitions by subtracting the initial energy from the final energy: (a) \( \Delta E_{(3 \rightarrow 4)} = E_4 - E_3 = \frac{-13.6}{16} - \frac{-13.6}{9} \ \text{eV} \) (b) \( \Delta E_{(2 \rightarrow 1)} = E_1 - E_2 = -13.6 - \frac{-13.6}{4} \ \text{eV} \) (c) \( \Delta E_{(3 \rightarrow \infty)} = E_\infty - E_3 = 0 - \frac{-13.6}{9} \ \text{eV} \)
03

Simplify the expressions to find the energy changes

We can simplify the expressions calculated in the previous step to find the energy changes for each transition: (a) \( \Delta E_{(3 \rightarrow 4)} = \frac{13.6}{16} - \frac{13.6}{9} = \frac{13.6 \times (9 - 16)}{144} = \frac{13.6 \times (-7)}{144} = -0.66 \ \text{eV} \) (b) \( \Delta E_{(2 \rightarrow 1)} = -13.6 + \frac{13.6}{4} = -13.6 \times (1 - \frac{1}{4}) = -13.6 \times \frac{3}{4} = -10.2 \ \text{eV} \) (c) \( \Delta E_{(3 \rightarrow \infty)} = 0 + \frac{13.6}{9} = \frac{13.6}{9} = 1.51 \ \text{eV} \) So, the energy changes corresponding to the transitions of the hydrogen atom are: (a) from n=3 to n=4: -0.66 eV (b) from n=2 to n=1: -10.2 eV (c) from n=3 to n=∞: 1.51 eV

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