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Chapter 7: Problem 72

Consider an electron for a hydrogen atom in an excited state. The maximum wavelength of electromagnetic radiation that can completely remove (ionize) the electron from the H atom is 1460 \(\mathrm{nm}\) . What is the initial excited state for the electron \((n=?) ?\)

Short Answer

Expert verified
The initial excited state for the electron in the hydrogen atom is \(n=2\).

Step by step solution

01

Recall the Rydberg formula for hydrogen atoms

The Rydberg formula for hydrogen atoms is given by: \[\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\] where - \(\lambda\) is the wavelength of the emitted radiation - \(R_H\) is the Rydberg constant for hydrogen (\(1.097 \times 10^7 \mathrm{m}^{-1}\)) - \(n_1\) and \(n_2\) are the initial and final principal quantum numbers, respectively, with \(n_1 < n_2\) In this case, the electron is being completely ionized, meaning the final energy level is infinite. Thus, we consider \(n_2 = \infty\).
02

Simplify the Rydberg formula

Since \(n_2 = \infty\), the term \(\frac{1}{n_2^2}\) approaches zero. Therefore, the Rydberg formula simplifies to: \[\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} \right)\]
03

Solve for the initial principal quantum number

Now we can rearrange the equation and solve for the initial principal quantum number (\(n_1\)). First, isolate \(n_1^2\): \[n_1^2 = \frac{R_H}{\frac{1}{\lambda}}\] Next, plug in the given values for \(\lambda\) and \(R_H\): \[n_1^2 = \frac{1.097 \times 10^7 \mathrm{m}^{-1}}{\frac{1}{1460 \times 10^{-9} \mathrm{m}}}\]
04

Calculate the value of n

Calculate the value of \(n_1\): \[n_1 = \sqrt{\frac{1.097 \times 10^7 \mathrm{m}^{-1}}{\frac{1}{1460 \times 10^{-9} \mathrm{m}}}}\] \[n_1 ≈ \sqrt{2.941}\] \(n_1 ≈ 1.71\) Since the principal quantum number must be a whole number, we round up to the nearest whole number. \(n_1 = 2\) Thus, the initial excited state for the electron in the hydrogen atom is \(n=2\).

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