Problem 144 - Chemistry

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.

Step-by-Step Solution

Answer

The required energy to remove an electron from the \(n=1\) state is \(19.62 x 10^{-18} J\), from the \(n=5\) state it is \(0.251 x 10^{-18} J\). The energy of the emitted photon during the transition from \(n=5\) to \(n=1\) is \(120.91 eV\), and the wavelength of this photon is \(102.67 nm\).

Step 1: Calculating the Ionization Energy for the \(n=1\) State

The ionization energy for hydrogen-like atoms is calculated using the formula \(E_n = -RZ^2/n^2 \), where \(E_n\) is the energy at a given state \(n\), \(R\) is the Rydberg constant for hydrogen like ions (\(2.18 x 10^{-18} J\)), \(Z\) is the atomic number, and \(n\) is the electron state. For the lithium ion \(\mathrm{Li}^{2+}\), \(Z=3\) and for the first state \(n=1\). Substituting these values into the formula, we find \(E_1 = -RZ^2 = -2.18 x 10^{-18} J(3)^2= -19.62 x 10^{-18} J\). The energy is negative, indicating that energy is needed to clear electron from the state, that's why it's multiplied by \(-1\). The ionization energy for the state \(n=1\) is therefore \(19.62 x 10^{-18}J\).

Step 2: Calculating the Ionization Energy for the \(n=5\) State

We use the same formula as in Step 1 but replace \(n\) with \(5\). The equation hence becomes \(E_5 = -RZ^2/5^2 = -2.18 x 10^{-18} J(3)^2/5^2= -0.251 x 10^{-18} J\). Multiplying by \(-1\), the ionization energy for the state \(n=5\) is \(0.251 x 10^{-18} J\).

Step 3: Calculation of Transition Energy and Conversion to electronvolt

The energy of the photon emitted during the transition from \(n=5\) to \(n=1\) is the difference between the energies at these levels. So, we subtract the energy at \(n=5\) (found in step 2) from the energy at \(n=1\) (found in step 1). The result is \(E_{transition} = E_1 - E_5 = 19.62 x 10^{-18} J - 0.251 x 10^{-18} J = 19.369 x 10^{-18} J\). To convert this energy to electronvolt (eV), we use the conversion factor of \(1eV = 1.60218 x 10^{-19} J\). Hence, \(E_{transition} (eV) = 19.369 x 10^{-18} J/(1.60218 x 10^{-19} J) = 120.91 eV\).

Step 4: Calculating the Wavelength of the Emitted Photon

We can apply the energy-wavelength relation \(E = hc/\lambda\), where \(E\) is the energy, \(h\) is the Planck's constant (\(6.626 x 10^{-34} Js\)), \(c\) is the speed of light (\(3.0 x 10^{8} ms^{-1}\)), and \(\lambda\) is the wavelength. Hence, \(\lambda = hc/E\). Substituting the values, \(\lambda = 6.626 x 10^{-34} Js*3.0 x 10^{8} ms^{-1}/19.369 x 10^{-18} J = 102.67 x 10^{-9} m\). We need to convert this to nanometers (nm) because the question calls for the answer in that unit. Knowing that \(1m = 10^9 nm\), we multiply by \(10^9\) to get \(\lambda = 102.67 nm\).

Key Concepts

Rydberg Constant for Ions

The Rydberg constant for ions is an essential element in quantum mechanics, specifically when dealing with the energy levels of hydrogen-like ions, which are atoms that have lost all but one of their electrons. When calculating the ionization energy required to remove this single electron, the Rydberg formula is used. The standard Rydberg constant value for a hydrogen atom is approximately 2.18 x 10^-18 J. However, for ions, the constant must be adjusted to account for the increased nuclear charge, achieved by multiplying the Rydberg constant by the square of the atomic number (Z²).

In the context of our exercise, the lithium ion (Li²⁺), having an atomic number of Z=3, modifies the Rydberg constant to be used in the energy calculation formula as:
\( R_{ion} = R\times Z^{2} \).
This adjustment of the Rydberg constant for ions ensures that we accurately calculate the energy levels for ions that differ significantly from neutral hydrogen atoms due to the much stronger electrostatic attraction between the nucleus and the remaining electron.

Electron Transition Energy

Understanding electron transition energy is crucial when dealing with phenomena such as emission of light or ionization processes. Transition energy refers to the energy difference between two electron orbits, or energy levels, in an atom. In hydrogen-like ions, an electron can transition from a higher energy level to a lower one, releasing energy in the form of a photon.

Each energy level in an atom is quantized and defined by the principal quantum number (n). The transition energy, in this case between the states from n=5 to n=1, is calculated using the formula:\[ E_{transition} = E_{n=1} - E_{n=5} \],where \( E_{n} \) is the energy of the electron in the nth state. In our exercise, the sizable difference in energy between these two states corresponds to the emission of a photon with considerable energy. The value obtained after calculation reveals the precise energy of the photon that is released when the electron makes this specific transition. This is vital to understand because the energy of the photon directly determines the color of the light observed, with high-energy transitions corresponding to light towards the blue end of the spectrum, while lower-energy transitions yield light towards the red end.

Wavelength of Emitted Photon

The wavelength of an emitted photon is inversely related to its energy, a fundamental concept in the field of spectroscopy. When an electron transitions between energy levels, the energy of the photon that is emitted during this process can be converted into a wavelength using the relationship:\[ E = \frac{hc}{\lambda} \],where E is the photon’s energy, h is the Planck constant, c is the speed of light, and \( \lambda \) is the wavelength. This is also known as the energy-wavelength relation.

In our specific exercise, we calculated the transition energy in joules and then expressed it in the more practical unit of electron volts (eV). To find the photon's wavelength in nanometers (nm), we re-arranged the energy-wavelength equation and solved for \( \lambda \). Such a photon wavelength corresponds to the color of light seen when the electron shifts from a higher energy level to a lower one. In spectroscopy, this information is used to identify elements and ions, as the wavelength of light emitted or absorbed is like a fingerprint, unique to each element or ion.