Question

Calculate the energies needed to remove an electron from the \(n=1\) state and the \(n=5\) state in the \(\mathrm{Li}^{2+}\) ion. What is the wavelength (in \(\mathrm{nm}\) ) of the emitted photon in a transition from \(n=5\) to \(n=1 ?\) The Rydberg constant for hydrogen-like ions is \((2.18 \times\) \(\left.10^{-18} \mathrm{~J}\right) Z^{2},\) where \(Z\) is the atomic number.

Short answer

The energy required to remove an electron from \(n = 1\) in \(\mathrm{Li}^{2+}\) is -122.4 eV, and from \(n = 5\) is -7.392 eV. The wavelength of the emitted photon in a transition from \(n = 5\) to \(n = 1\) is 108 nm.

Step-by-step solution

Calculate the Energy for n=1

Use the formula \(-13.6 \times \frac{Z^{2}}{n^{2}}\) eV to get \(-13.6 \times \frac{3^{2}}{1^{2}}\) eV = -122.4 eV.

Calculate the Energy for n=5

Similarly, substitute \(Z = 3\) and \(n = 5\) into the formula, getting \(-13.6 \times \frac{3^{2}}{5^{2}}\) eV = -7.392 eV.

Calculate the Energy Difference ∆E

Subtract the energy at level n=5 from the energy at level n=1 to calculate the energy difference: ∆E = -122.4 eV - (-7.392 eV) = -115.008 eV. Convert the energy difference from eV to Joules by multiplying by 1.6×10^-19 J/eV, giving ∆E = -115.008 eV × 1.6×10^-19 J/eV = -1.84×10^-17 J.

Calculate the Wavelength λ of the Emitted Photon

Use the formula E=hc/λ, where h is Planck's constant (h = 6.626×10^-34 J∙s), c is the speed of light (c = 3×10^8 m/s), and E is the energy of the transition from step 3. Solving for λ gives λ = hc/E = (6.626×10^-34 J∙s × 3×10^8 m/s) / -1.84×10^-17 J = -1.08×10^-7 m. Convert the wavelength from meters to nanometers by multiplying by 10^9 nm/m, giving λ = -1.08×10^-7 m × 10^9 nm/m = -108 nm. The wavelength cannot be negative, so the absolute value is taken, yielding λ = 108 nm.

Key concepts

Electron Transition

Electron transitions are changes in the energy state of an electron within an atom or ion. When an electron moves from a higher energy level (like from the 5th orbital) to a lower one (such as the 1st orbital), energy is released. In contrast, absorbing energy may promote an electron to a higher energy level.
This process of moving between energy levels is crucial in the study of atomic spectra as it helps determine the energy and wavelength of photons involved. For example, in our exercise, an electron transitions from the n=5 state to the n=1 state in a lithium ion, releasing energy in the form of emitted photons.
Breaking down these transitions and their resulting photon emissions requires understanding several concepts in atomic physics.

Hydrogen-like Ions

Hydrogen-like ions refer to any ion that has only one electron remaining, similar to a hydrogen atom. Although the most common example of this is the hydrogen atom itself, any atom that loses all but one of its electrons becomes hydrogen-like. An example is the \(\text{Li}^{2+}\) ion, which has only one electron.
These ions are much easier to study because their atomic structure is simpler, allowing us to apply concepts like the Rydberg formula that describe electron transitions. In hydrogen-like ions, knowing the atomic number (Z) is essential. It reflects the number of protons in the nucleus and influences the energy levels of the electrons orbiting the nucleus. Thus, calculations of energy transitions in such ions often involve Z, further affecting the energy changes during electron transitions.

Wavelength Calculation

The wavelength of the photon emitted or absorbed during an electron transition is found using the relationship between energy and wavelength. The formula used is \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) (lambda) is the wavelength.
In the context of our exercise, after determining the energy difference when transitioning from \(n=5\) to \(n=1\) (noting it is 1.84 × 10^-17 J and taking the absolute value), you can solve for \(\lambda\). Rearrange the formula to obtain: \(\lambda = \frac{hc}{|\Delta E|}\).
Calculating the wavelength provides insight into the nature of the light emitted, with shorter wavelengths indicating higher energy radiation, such as ultraviolet light.

Planck's Constant

Planck's constant (\(h\=6.626 × 10^{-34} \Js\) plays a pivotal role in quantum mechanics, serving as a bridge between the energy of a photon and the frequency of its associated electromagnetic wave.
It's especially crucial when working with energy and wavelength calculations, alongside the speed of light (\(c = 3×10^8 \frac{m}{s}\).In this context, Planck's constant is used to connect the energy change (\(\Delta E\)) to the photon wavelength (\(\lambda\)), using the equation \(E = \frac{hc}{\lambda}\).Understanding Planck's constant allows one to appreciate the quantized nature of energy transitions, where these constants are instrumental in converting physical changes to measurable wavelengths.

Atomic Number

The atomic number, symbolized as \(Z\), represents the number of protons present in the nucleus of an atom and defines the chemical element's identity.
While calculating energy levels and transition energies in hydrogen-like ions, \(Z\) is paramount. In equations involving energy levels, the \(Z^2\) term significantly impacts the magnitude of calculated energies. For instance, in our exercise with the \(\mathrm{Li}^{2+}\) ion, the atomic number is 3.

• The higher the atomic number, the stronger the positive charge of the nucleus, which in turn affects the energy levels.
• This increased nuclear attraction lowers the energy of the electron orbits.

Thus, energies needed to remove electrons from these orbits are higher, and the shifts play a crucial role in determining the wavelength of light or other radiation emitted during transitions.