Radiation from a helium ion He+ is nearly equal in wavelength to the Hα line (the first line of the Balmer series). The states (values of n) between which the transition in the helium ion occur are
Helium Ion Radiation and H-alpha Line
The question asks for the states (principal quantum numbers, n) involved in a transition in a helium ion (He+) that emits radiation with a wavelength nearly equal to the Hα line of hydrogen.
Both hydrogen and helium ion (He+) are hydrogen-like species, meaning they have only one electron orbiting the nucleus. The wavelengths of emitted radiation during electron transitions in such species can be calculated using the Rydberg formula.
Rydberg Formula for Wavelength
The formula for the reciprocal wavelength (\(1/\lambda\)) of radiation emitted during a transition from an initial state with principal quantum number \(n_i\) to a final state with principal quantum number \(n_f\) (\(n_i > n_f\)) in a hydrogen-like atom or ion with atomic number Z is given by:
\(\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\)
where R is the Rydberg constant.
H-alpha Line Calculation
The Hα line is the first line of the Balmer series in the hydrogen spectrum. For hydrogen, the atomic number Z=1. The Balmer series corresponds to transitions where the electron falls to the \(n_f = 2\) energy level. The first line of this series is the transition from the next higher energy level, which is \(n_i = 3\), down to \(n_f = 2\).
Using the Rydberg formula for the Hα line (Z=1, \(n_i=3\), \(n_f=2\)):
\(\frac{1}{\lambda_{H\alpha}} = R (1)^2 \left(\frac{1}{2^2} - \frac{1}{3^2}\right)\)
\(\frac{1}{\lambda_{H\alpha}} = R \left(\frac{1}{4} - \frac{1}{9}\right)\)
\(\frac{1}{\lambda_{H\alpha}} = R \left(\frac{9 - 4}{36}\right)\)
\(\frac{1}{\lambda_{H\alpha}} = R \left(\frac{5}{36}\right)\)
Helium Ion Transition Analysis
For the helium ion (He+), the atomic number is Z=2. We are looking for a transition between states \(n_i\) and \(n_f\) (\(n_i > n_f\)) such that the emitted radiation has a wavelength nearly equal to the Hα line. This means their reciprocal wavelengths are nearly equal:
\(\frac{1}{\lambda_{He^+}} \approx \frac{1}{\lambda_{H\alpha}}\)
Using the Rydberg formula for He+ (Z=2, transition from \(n_i\) to \(n_f\)):
\(\frac{1}{\lambda_{He^+}} = R (2)^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\)
\(\frac{1}{\lambda_{He^+}} = 4R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\)
Equating the reciprocal wavelengths (approximately):
\(4R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \approx R \left(\frac{5}{36}\right)\)
Dividing by R, we get the condition for the transition in He+:
\(4 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) \approx \frac{5}{36}\)
Checking Options for He+ Transitions
We examine the given options for the initial and final states (\(n_i, n_f\)) and calculate the value of \(4 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\) for each transition (assuming \(n_i > n_f\) for emission). The target value for \(4 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\) is \(\frac{5}{36}\).
| Option | States (\(n_i, n_f\)) | \(\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\) | \(4 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\) | Comparison to \(\frac{5}{36}\) |
|---|---|---|---|---|
| 1 | (6, 2) | \(\frac{1}{2^2} - \frac{1}{6^2} = \frac{1}{4} - \frac{1}{36} = \frac{9-1}{36} = \frac{8}{36} = \frac{2}{9}\) | \(4 \times \frac{2}{9} = \frac{8}{9}\) | \(\frac{8}{9} = \frac{32}{36}\) (Not \(\approx \frac{5}{36}\)) |
| 2 | (3, 1) | \(\frac{1}{1^2} - \frac{1}{3^2} = \frac{1}{1} - \frac{1}{9} = \frac{9-1}{9} = \frac{8}{9}\) | \(4 \times \frac{8}{9} = \frac{32}{9}\) | \(\frac{32}{9} = \frac{128}{36}\) (Not \(\approx \frac{5}{36}\)) |
| 3 | (3, 2) | \(\frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}\) | \(4 \times \frac{5}{36} = \frac{5}{9}\) | \(\frac{5}{9} = \frac{20}{36}\) (Not \(\approx \frac{5}{36}\)) |
| 4 | (6, 4) | \(\frac{1}{4^2} - \frac{1}{6^2} = \frac{1}{16} - \frac{1}{36} = \frac{9-4}{144} = \frac{5}{144}\) | \(4 \times \frac{5}{144} = \frac{20}{144} = \frac{5}{36}\) | Exactly equal to \(\frac{5}{36}\) |
Comparison and Conclusion
Comparing the calculated values to the required value of \(\frac{5}{36}\), we find that the transition from n=6 to n=4 in He+ yields radiation with a reciprocal wavelength exactly equal to that of the Hα line in Hydrogen.
The provided correct answer specifies the states as (3, 1). For the transition from n=3 to n=1 in He+, the value of \(4 \left(\frac{1}{1^2} - \frac{1}{3^2}\right)\) is \(\frac{32}{9}\). Based on the standard Rydberg formula calculation, this value is significantly different from \(\frac{5}{36}\).
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