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Chapter 2: Problem 61

Calculate the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the \(n=6\) state and then fall to states with smaller values of \(n\).

Short Answer

Expert verified
The longest wavelength of light emitted is approximately \(8.10 \times 10^{-7}\) m when the electron transitions from n=6 to n=5, while the shortest wavelength of light emitted is approximately \(9.69 \times 10^{-8}\) m when the electron transitions from n=6 to n=1.

Step by step solution

01

Determine the Transition for the Longest Wavelength

We need to find the smallest energy difference among all possible transitions. The smallest energy difference occurs when the electron moves from n=6 (the initial state) to n=5 (the next smallest quantum number).
02

Calculate the Longest Wavelength

Now we will use the Rydberg formula with n1 = 6 and n2 = 5: \[\frac{1}{\lambda_{long}} = R_H \left(\frac{1}{5^2} - \frac{1}{6^2}\right)\] Plug in the Rydberg constant value: \[\frac{1}{\lambda_{long}} = (1.097 \times 10^7 \,\text{m}^{-1}) \left(\frac{1}{25} - \frac{1}{36}\right)\] Calculate the reciprocal of the wavelength: \[\frac{1}{\lambda_{long}} = 1.235 \times 10^6 \,\text{m}^{-1}\] Now find the longest wavelength by taking the reciprocal: \[\lambda_{long} = \frac{1}{1.235 \times 10^6 \,\text{m}^{-1}} \approx 8.10 \times 10^{-7} \,\text{m}\]
03

Determine the Transition for the Shortest Wavelength

We need to find the largest energy difference among all possible transitions. The largest energy difference occurs when the electron moves from n=6 (the initial state) to n=1 (the smallest quantum number).
04

Calculate the Shortest Wavelength

Now we will use the Rydberg formula with n1 = 6 and n2 = 1: \[\frac{1}{\lambda_{short}} = R_H \left(\frac{1}{1^2} - \frac{1}{6^2}\right)\] Plug in the Rydberg constant value: \[\frac{1}{\lambda_{short}} = (1.097 \times 10^7 \,\text{m}^{-1}) \left(\frac{1}{1} - \frac{1}{36}\right)\] Calculate the reciprocal of the wavelength: \[\frac{1}{\lambda_{short}} = 1.032 \times 10^7 \, \text{m}^{-1}\] Now find the shortest wavelength by taking the reciprocal: \[\lambda_{short} = \frac{1}{1.032 \times 10^7 \,\text{m}^{-1}} \approx 9.69 \times 10^{-8} \,\text{m}\]
05

Final Answer

The longest wavelength of light emitted is approximately \(8.10 \times 10^{-7}\) m, while the shortest wavelength of light emitted is approximately \(9.69 \times 10^{-8}\) m.

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