This question involves calculating the wavelength of light corresponding to an electronic transition in 1,3,5-hexatriene, which is modeled using the particle in a one-dimensional box concept.
Particle in a 1D Box Model for Hexatriene
The 1,3,5-hexatriene molecule has a conjugated system with 3 double bonds. This system contains $\pi$ electrons that can be treated as particles in a 1D box.
- Number of $\pi$ electrons ($n_e$): 1,3,5-hexatriene has 3 double bonds, so it has $n_e = 6$ $\pi$ electrons.
- Energy Levels: In the particle in a 1D box model, the energy levels are quantized and given by the formula:
$E_n = \frac{n^2 h^2}{8mL^2}$
where $n$ is the principal quantum number ($n=1, 2, 3, \dots$), $h$ is Planck's constant, $m$ is the mass of the electron, and $L$ is the length of the box. - Electron Configuration: The 6 $\pi$ electrons fill the lowest available energy levels. According to the Pauli exclusion principle, each energy level can hold a maximum of 2 electrons. Therefore, the electrons occupy the levels $n=1$, $n=2$, and $n=3$.
- Transition: The transition from the ground state to the first excited state involves promoting an electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). In this case, the HOMO is $n=3$ and the LUMO is $n=4$. The transition is from $n_{initial}=3$ to $n_{final}=4$.
Calculating Energy Difference ($\Delta E$)
The energy difference for the $n=3 \to n=4$ transition is calculated as:
$\Delta E = E_4 - E_3 = \frac{4^2 h^2}{8mL^2} - \frac{3^2 h^2}{8mL^2}$
Factoring out the common term $\frac{h^2}{8mL^2}$:
$\Delta E = \frac{h^2}{8mL^2} (4^2 - 3^2) = \frac{h^2}{8mL^2} (16 - 9) = 7 \frac{h^2}{8mL^2}$
Determining Wavelength ($\lambda$)
The energy absorbed corresponds to the energy of a photon ($\Delta E = h\nu = \frac{hc}{\lambda}$). We can find the wavelength using this relationship:
$\lambda = \frac{hc}{\Delta E}$
Substituting the expression for $\Delta E$:
$\lambda = \frac{hc}{7 \frac{h^2}{8mL^2}} = \frac{8mL^2 c}{7h}$
Numerical Calculation Steps
Substitute the known values and constants:
- Box length, $L = 0.70$ nm $= 0.70 \times 10^{-9}$ m
- Planck's constant, $h = 6.626 \times 10^{-34}$ J·s
- Mass of electron, $m = 9.109 \times 10^{-31}$ kg
- Speed of light, $c = 3.00 \times 10^8$ m/s
Calculate intermediate values:
- $L^2 = (0.70 \times 10^{-9} \text{ m})^2 = 0.49 \times 10^{-18} \text{ m}^2$
- $8mL^2 = 8 \times (9.109 \times 10^{-31} \text{ kg}) \times (0.49 \times 10^{-18} \text{ m}^2) \approx 3.571 \times 10^{-48} \text{ kg m}^2$
- $h^2 = (6.626 \times 10^{-34} \text{ J·s})^2 \approx 4.390 \times 10^{-67} \text{ J}^2\text{s}^2$
Calculate the energy difference $\Delta E$:
$\Delta E = 7 \times \frac{h^2}{8mL^2} = 7 \times \frac{4.390 \times 10^{-67} \text{ J}^2\text{s}^2}{3.571 \times 10^{-48} \text{ kg m}^2}$
Note: $1 \text{ J} = 1 \text{ kg m}^2/\text{s}^2$, so $1 \text{ J}^2\text{s}^2 / (\text{kg m}^2) = 1 \text{ J}$
$\Delta E \approx 7 \times (1.229 \times 10^{-19} \text{ J}) \approx 8.603 \times 10^{-19}$ J
Calculate the wavelength $\lambda$:
$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \text{ J·s}) \times (3.00 \times 10^8 \text{ m/s})}{8.603 \times 10^{-19} \text{ J}}$
$\lambda \approx \frac{1.9878 \times 10^{-25}}{8.603 \times 10^{-19}} \text{ m} \approx 2.311 \times 10^{-7}$ m
Convert the wavelength to nanometers:
$\lambda = 2.311 \times 10^{-7} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} = 231.1$ nm
The calculated wavelength of 231.1 nm falls within the expected range of 225 nm to 240 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)
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 $m_e =9.1\times10^{-31} \text{ kg}$, $h=6.6\times10^{-34} \text{ Js}$, $c=3.0\times10^8 \text{ m s}^{-1}$)
