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Chapter 7: Problem 73

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{s}^{-1} .\) Determine the principal quantum level for the final state in this electronic transition.

Short Answer

Expert verified
The final principal quantum level in this electronic transition is \(n_f = 4\).

Step by step solution

01

Rydberg Formula

The Rydberg formula for the hydrogen atom is given by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \] where \(R_H\) is the Rydberg constant, \(n_i\) and \(n_f\) represent the initial and final states, respectively. To solve for the final state n_f, we will first convert the given frequency to the wavelength and then use the formula to find \(n_f\). 2. Calculate the wavelength from the given frequency
02

Wavelength Calculation

The relationship between frequency (\(v\)), wavelength (\(\lambda\)) and the speed of light (\(c\)) is given as: \[ \lambda = \frac{c}{v} \] Given the frequency \(v = 6.90 \times 10^{14} \mathrm{s}^{-1}\) and the speed of light \(c = 3.00 \times 10^8 \mathrm{m}\cdot\mathrm{s}^{-1}\), the wavelength can be calculated as: \[ \lambda = \frac{3.00 \times 10^8 \mathrm{m}\cdot\mathrm{s}^{-1}}{6.90 \times 10^{14} \mathrm{s}^{-1}} = 4.35 \times 10^{-7} \mathrm{m} \] 3. Substitute values in the Rydberg formula.
03

Substituting Values

Our known values are \(\lambda = 4.35 \times 10^{-7} \mathrm{m}\), the Rydberg constant \(R_H = 1.097 \times 10^7 \mathrm{m}^{-1}\), and the initial quantum level \(n_i=5\). Substitute these values into the Rydberg formula: \[ \frac{1}{4.35 \times 10^{-7} \mathrm{m}} = 1.097 \times 10^7 \mathrm{m}^{-1} \left( \frac{1}{5^2} - \frac{1}{n_f^2} \right) \] 4. Solve for the final quantum level
04

Solving for Final Quantum Level

Re-write the formula to isolate \(n_f\): \[ \frac{1}{n_f^2} = \frac{1}{5^2} - \frac{1}{4.35 \times 10^{-7} \mathrm{m} \times 1.097 \times 10^7 \mathrm{m}^{-1}} \] Calculate \(\frac{1}{n_f^2}\): \[ \frac{1}{n_f^2} = \frac{1}{25} - \frac{1}{4.761 \times 10^{-7}} \] Solve for \(n_f\): \[ n_f^2 = \frac{1}{(\frac{1}{25} - \frac{1}{4.761 \times 10^{-7}})^{-1}} \] \[ n_f^2 \approx 15.97 \] \[ n_f = \sqrt{15.97} \approx 3.99 \] 5. Conclusion
05

Final Answer

As the quantum levels have to be whole numbers and the square root found is closer to 4, the final principal quantum level in this electronic transition is \(n_f = 4\).

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