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Chapter 7: Problem 106

The \(\mathrm{He}^{+}\) ion contains only one electron and is therefore a hydrogen-like ion. Calculate the wavelengths, in increasing order, of the first four transitions in the Balmer series of the \(\mathrm{He}^{+}\) ion. Compare these wavelengths with the same transitions in a \(\mathrm{H}\) atom. Comment on the differences. (The Rydberg constant for \(\mathrm{He}^{+}\) is \(\left.8.72 \times 10^{-18} \mathrm{~J} .\right)\)

Short Answer

Expert verified
The wavelengths of the first four transitions in the Balmer series of the \( \mathrm{He}^{+} \) ion are shorter than those in the hydrogen atom due to the higher Rydberg constant of the \( \mathrm{He}^{+} \) ion, indicating higher energy changes in transitions.

Step by step solution

01

Title: Calculate the Wavelength of Transition in \( \mathrm{He}^{+} \) Ion

We know that the wavelength for any transition can be calculated using Rydberg's formula: \[ \frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right) \] Where \( R \) is the Rydberg constant, \( n_{1} \) is the lower energy level and \( n_{2} \) is the higher energy level. For Balmer series, \( n_{1} = 2 \). The first four transitions correspond to \( n_{2} \) values of 3, 4, 5 and 6. We can substitute these values into the Rydberg formula along with the given Rydberg constant for \( \mathrm{He}^{+} \), then we solve for \( \lambda \) (wavelength).
02

Title: Calculate the Wavelength of Transition in the Hydrogen Atom

Next, we do the same for a hydrogen atom. The Rydberg constant for an hydrogen atom \( H \) is \( 1.097373 \times 10^{7} \mathrm{ m^{-1}} \). Like earlier, we substitute \( n_{1} = 2 \), the first four \( n_{2} \) values of 3, 4, 5 and 6 into Rydberg's formula and solve for \( \lambda \).
03

Title: Compare the Wavelengths and Comment on the Differences

We compare the calculated wavelengths of the \( \mathrm{He}^{+} \) and \( \mathrm{H} \) atoms. Note that because the \( \mathrm{He}^{+} \) ion has a larger Rydberg constant than the hydrogen atom, the wavelengths of its transitions will generally be shorter, which means the transitions involve a larger energy change. These differences in transition energies are what make helium and other elements distinct from hydrogen in their spectroscopic characteristics.

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Most popular questions from this chapter

The retina of a human eye can detect light when radiant energy incident on it is at least \(4.0 \times 10^{-17} \mathrm{~J}\) For light of 600 -nm wavelength, how many photons does this correspond to?

Explain why elements produce their own characteristic colors when they emit photons.

Why is a boundary surface diagram useful in representing an atomic orbital?

When an electron makes a transition between energy levels of a hydrogen atom, there are no restrictions on the initial and final values of the principal quantum number \(n\). However, there is a quantum mechanical rule that restricts the initial and final values of the orbital angular momentum \(\ell\). This is the selection rule, which states that \(\Delta \ell=\pm 1,\) that is, in a transition, the value of \(\ell\) can only increase or decrease by one. According to this rule, which of the following transitions are allowed: (a) \(1 s \longrightarrow 2 s\), (b) \(2 p \longrightarrow 1 s\) (c) \(1 s \longrightarrow 3 d\) (d) \(3 d \longrightarrow 4 f\), (e) \(4 d \longrightarrow 3 s ?\)

List all the possible subshells and orbitals associated with the principal quantum number \(n\), if \(n=5\).
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