The $\pi$ electrons in benzene can be modelled as particles in a ring that follow Pauli's exclusion principle. Given that the radius of benzene is 1.4 Å, the longest wavelength of light that is absorbed during an electronic transition in benzene is ____________ nm. (Up to one decimal place. Use $m_e =9.1\times10^{-31} \text{ kg}$, $h=6.6\times10^{-34} \text{ Js}$, $c=3.0\times10^8 \text{ m s}^{-1}$)
Benzene Pi Electron Wavelength Calculation
This problem requires calculating the longest wavelength of light absorbed by benzene's $\pi$ electrons using the particle in a ring model. This longest wavelength corresponds to the lowest energy electronic transition, typically from the Highest Occupied Molecular Orbital (HOMO) to the Lowest Unoccupied Molecular Orbital (LUMO).
Particle in a Ring Model for Benzene
Benzene has 6 $\pi$ electrons. These electrons are treated as particles moving in a ring of radius $r$. The energy levels for a particle in a ring are given by:
$E_n = \frac{n^2 h^2}{2mL^2}$where $n$ is the quantum number ($n = 0, \pm 1, \pm 2, \dots$), $h$ is Planck's constant, $m$ is the mass of the electron, and $L$ is the circumference of the ring ($L = 2\pi r$).
Electronic Configuration and Transition
According to Pauli's exclusion principle, each energy level can hold a maximum of two electrons with opposite spins. For benzene's 6 $\pi$ electrons, the energy levels are filled as follows:
- $n=0$: 2 electrons
- $n=\pm 1$: 4 electrons (2 for $n=1$, 2 for $n=-1$)
The ground state configuration is $(n=0)^2 (n=\pm 1)^4$. The HOMO corresponds to the $n=\pm 1$ levels, and the LUMO corresponds to the next available levels, $n=\pm 2$. The transition responsible for the longest absorbed wavelength is from HOMO ($n=1$ or $n=-1$) to LUMO ($n=2$ or $n=-2$).
The energy difference ($\Delta E$) for this transition is:
$ \Delta E = E_2 - E_1 = \frac{2^2 h^2}{2mL^2} - \frac{1^2 h^2}{2mL^2} = \frac{(4-1)h^2}{2mL^2} = \frac{3h^2}{2mL^2} $Wavelength Calculation
The energy of absorbed light is related to its wavelength ($\lambda$) by $\Delta E = \frac{hc}{\lambda}$. Therefore, the wavelength is:
$ \lambda = \frac{hc}{\Delta E} = \frac{hc}{\frac{3h^2}{2mL^2}} = \frac{2mL^2 c}{3h} $Substituting $L = 2\pi r$:
$ \lambda = \frac{2m(2\pi r)^2 c}{3h} = \frac{2m \cdot 4\pi^2 r^2 c}{3h} = \frac{8\pi^2 m c r^2}{3h} $Applying Given Values
Given:
- Radius, $r = 1.4 \text{ Å} = 1.4 \times 10^{-10} \text{ m}$
- Electron mass, $m = 9.1 \times 10^{-31} \text{ kg}$
- Planck's constant, $h = 6.6 \times 10^{-34} \text{ Js}$
- Speed of light, $c = 3.0 \times 10^8 \text{ m s}^{-1}$
- Use $\pi \approx 3.14159$
Plugging these values into the formula for $\lambda$:
$ \lambda = \frac{8 \times (3.14159)^2 \times (9.1 \times 10^{-31} \text{ kg}) \times (3.0 \times 10^8 \text{ m s}^{-1}) \times (1.4 \times 10^{-10} \text{ m})^2}{3 \times (6.6 \times 10^{-34} \text{ Js})} $Calculate the numerator:
Numerator $\approx 8 \times 9.8696 \times 9.1 \times 10^{-31} \times 3.0 \times 10^8 \times 1.96 \times 10^{-20}$
Numerator $\approx 4224.82 \times 10^{-43} \text{ kg m}^3 \text{ s}^{-1}$
Calculate the denominator:
Denominator $= 3 \times 6.6 \times 10^{-34} = 19.8 \times 10^{-34} \text{ Js}$
Calculate wavelength:
$ \lambda = \frac{4224.82 \times 10^{-43}}{19.8 \times 10^{-34}} \text{ m} $ $ \lambda \approx 213.37 \times 10^{-9} \text{ m} $ $ \lambda \approx 213.4 \text{ nm} $The longest wavelength of light absorbed during an electronic transition in benzene is approximately 213.4 nm.
Important Questions from Particle in a Box
Consider two non-interacting particles confined to a one-dimensional box with infinite potential barriers. Their wavefunctions are $\psi_1$ and $\psi_2$ and energies are $E_1$ and $E_2$, respectively. The INCORRECT statement(s) about this system is/are
- Wavefunctions and energies for a particle confined in a cubic box are $\psi_{n_x,n_y,n_z}$ and $E_{n_x,n_y,n_z}$, respectively. The functions $\Phi_1$, $\Phi_2$, $\Phi_3$, and $\Phi_4$ are written as linear combinations of $\psi_{n_x,n_y,n_z}$. Among these functions, the eigenfunction(s) of the Hamiltonian operator for this particle is/are
$\Phi_1 = \frac{1}{\sqrt{2}}\psi_{1,4,1} - \frac{1}{\sqrt{2}}\psi_{2,2,3}$
$\Phi_2 = \frac{1}{\sqrt{2}}\psi_{1,5,1} + \frac{1}{\sqrt{2}}\psi_{3,3,3}$
$\Phi_3 = \frac{1}{\sqrt{2}}\psi_{1,3,8} + \frac{1}{\sqrt{2}}\psi_{3,8,1}$
$\Phi_4 = \frac{1}{2}\psi_{3,3,1} + \frac{\sqrt{3}}{2}\psi_{2,4,1}$ The wave function of a particle in a cubic box (of side L) is given by
$\psi(x, y, z) = \sqrt{32/L^3} \sin \frac{\pi x}{L} \cos \frac{\pi x}{L} \sin \frac{2\pi y}{L} \sin \frac{\pi z}{L}$.
The ratio of the energy of the state corresponding to the above wave function to the ground state energy is ________.
(rounded off to the nearest integer)- The wavelength associated with a particle in one-dimensional box of length $L$ is ($n$ refers to the quantum number)
- Assume 1,3,5-hexatriene to be a linear molecule and model the $\pi$ electrons as particles in a one-dimensional box of length 0.70 nm. The wavelength (in nm) corresponding to the transition from the ground-state to the first excited-state is ________
