Calculating Wavelength of He+ Transition

This problem asks us to calculate the wavelength of the radiation emitted when a He+ ion undergoes a transition from a higher energy level (n=3) to a lower energy level (n=2). This involves understanding electron transitions in hydrogenic atoms and using the appropriate formula.

He+ is a hydrogenic ion because it has only one electron orbiting the nucleus, similar to a hydrogen atom. For such systems, the energy levels and the wavelengths of emitted or absorbed radiation during transitions can be calculated using a modified version of the Rydberg formula for hydrogen.

Rydberg Formula for Hydrogenic Atoms

The formula relating the wavelength (λ) of emitted or absorbed radiation during an electron transition between two energy levels n\(_i\) (initial) and n\(_f\) (final) in a hydrogenic atom with atomic number Z is given by:

\(\frac{1}{\lambda} = R \cdot Z^2 \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\)

Where:

• \(\lambda\) is the wavelength of the emitted or absorbed radiation.
• \(R\) is the Rydberg constant.
• \(Z\) is the atomic number of the nucleus (number of protons).
• \(n_i\) is the principal quantum number of the initial energy level.
• \(n_f\) is the principal quantum number of the final energy level.

Applying the Formula to He+ Ion

For this specific problem involving the He+ ion making a transition from n=3 to n=2, we have the following values:

• Rydberg constant, \(R = 1.097 \times 10^7\) m⁻¹
• Atomic number of Helium, \(Z = 2\) (He+ has 2 protons)
• Initial energy level, \(n_i = 3\)
• Final energy level, \(n_f = 2\)

Now, let's substitute these values into the Rydberg formula:

\(\frac{1}{\lambda} = (1.097 \times 10^7 \text{ m}^{-1}) \cdot (2)^2 \left(\frac{1}{2^2} - \frac{1}{3^2}\right)\)

\(\frac{1}{\lambda} = (1.097 \times 10^7) \cdot 4 \left(\frac{1}{4} - \frac{1}{9}\right)\)

First, calculate the term inside the parenthesis:

\(\left(\frac{1}{4} - \frac{1}{9}\right) = \left(\frac{9}{36} - \frac{4}{36}\right) = \frac{5}{36}\)

Now substitute this back into the formula:

\(\frac{1}{\lambda} = (1.097 \times 10^7) \cdot 4 \cdot \left(\frac{5}{36}\right)\)

\(\frac{1}{\lambda} = (4.388 \times 10^7) \cdot \left(\frac{5}{36}\right)\)

\(\frac{1}{\lambda} = \frac{4.388 \times 10^7 \times 5}{36}\)

\(\frac{1}{\lambda} = \frac{21.94 \times 10^7}{36}\)

\(\frac{1}{\lambda} \approx 0.60944 \times 10^7 \text{ m}^{-1}\)

\(\frac{1}{\lambda} \approx 6.0944 \times 10^6 \text{ m}^{-1}\)

To find the wavelength \(\lambda\), we take the reciprocal of this value:

\(\lambda = \frac{1}{6.0944 \times 10^6 \text{ m}^{-1}}\)

\(\lambda \approx 0.16408 \times 10^{-6} \text{ m}\)

Converting Wavelength to Nanometers

The options for the wavelength are given in nanometers (nm) or micrometers (μm). Let's convert the wavelength from meters to nanometers. Remember that 1 meter = 10⁹ nanometers.

\(\lambda \approx 0.16408 \times 10^{-6} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}}\)

\(\lambda \approx 0.16408 \times 10^{(-6 + 9)} \text{ nm}\)

\(\lambda \approx 0.16408 \times 10^3 \text{ nm}\)

\(\lambda \approx 164.08 \text{ nm}\)

Rounding to one decimal place, we get 164.1 nm.

Let's also check in micrometers, just in case. 1 meter = 10⁶ micrometers.

\(\lambda \approx 0.16408 \times 10^{-6} \text{ m} \times \frac{10^6 \text{ }\mu\text{m}}{1 \text{ m}}\)

\(\lambda \approx 0.16408 \times 10^{(-6 + 6)} \text{ }\mu\text{m}\)

\(\lambda \approx 0.16408 \times 10^0 \text{ }\mu\text{m}\)

\(\lambda \approx 0.16408 \text{ }\mu\text{m}\)

Comparing our calculated value (164.1 nm) with the given options:

1. 1.641 nm
2. 164.1 nm
3. 16.41 μm (which is 16410 nm)
4. 1641 nm

Our calculated value of approximately 164.1 nm matches Option 2.

Summary of Wavelength Calculation

The calculation confirms that when a He+ ion transitions from the n=3 to the n=2 energy level, the emitted radiation has a wavelength close to 164.1 nm. This wavelength falls in the ultraviolet (UV) region of the electromagnetic spectrum.

Revision Table: Key Concepts for Atomic Transitions

Additional Information: Understanding Atomic Spectra and He+

The light emitted or absorbed by atoms and ions occurs at specific wavelengths, forming a unique spectrum. This is because electrons can only occupy discrete energy levels within the atom. When an electron jumps from one energy level to another, it emits or absorbs a photon with energy equal to the difference between the two levels. The energy of the photon is related to its wavelength by the equation \(E = \frac{hc}{\lambda}\), where h is Planck's constant and c is the speed of light.

The Rydberg formula is derived from Bohr's model of the atom, which successfully explained the spectrum of hydrogen. For hydrogenic ions like He+, the nuclear charge is different from hydrogen (\(Z=1\)). Since He+ has \(Z=2\), the electrostatic attraction between the nucleus and the electron is stronger than in hydrogen. This stronger attraction causes the energy levels in He+ to be lower (more negative) and spaced further apart than in hydrogen, scaled by \(Z^2\). Consequently, transitions in He+ involve larger energy differences and thus shorter wavelengths (or higher frequencies) compared to similar transitions in hydrogen.

For example, the transition from n=3 to n=2 in hydrogen corresponds to the first line of the Balmer series (H-alpha), which is in the visible red region (~656 nm). For He+, with \(Z=2\), the formula shows that the wavelength for the same transition (n=3 to n=2) will be shorter by a factor of \(1/Z^2 = 1/2^2 = 1/4\). So, the wavelength for He+ would be roughly 656 nm / 4 = 164 nm, which aligns with our calculated value and the answer option.