Particle in a 1D Box Energy Difference Calculation
The energy levels for a particle of mass m in a one-dimensional box of length L are given by the formula:
$ E_n = \frac{n^2 h^2}{8mL^2} $
where n is the principal quantum number (n=1 for the ground state), h is Planck's constant, and m is the mass of the particle.
Required Constants
- Planck's constant, h = 6.626 × 10-34 J·s
- Mass of electron, me = 9.109 × 10-31 kg
- Avogadro's number, NA = 6.022 × 1023 mol-1
Step-by-Step Solution
-
Calculate the energy difference per electron in Joules (J).
The ground state energy difference is:
$ \Delta E = E_{1, L_1} - E_{1, L_2} = \frac{h^2}{8m_e} \left( \frac{1}{L_1^2} - \frac{1}{L_2^2} \right) $
Given box lengths:
- L1 = 0.2 nm = 0.2 × 10-9 m
- L2 = 2 nm = 2 × 10-9 m
Calculate the squares of the lengths:
- L12 = (0.2 × 10-9 m)2 = 4 × 10-20 m2
- L22 = (2 × 10-9 m)2 = 4 × 10-18 m2
Calculate the constant factor:
$ \frac{h^2}{8m_e} = \frac{(6.626 \times 10^{-34} \text{ J·s})^2}{8 \times (9.109 \times 10^{-31} \text{ kg})} \approx 6.024 \times 10^{-38} \text{ J·m}^2 $
Substitute values to find the energy difference:
$ \Delta E = (6.024 \times 10^{-38} \text{ J·m}^2) \left( \frac{1}{4 \times 10^{-20} \text{ m}^2} - \frac{1}{4 \times 10^{-18} \text{ m}^2} \right) $
$ \Delta E = (6.024 \times 10^{-38}) \times (0.25 \times 10^{20} - 0.25 \times 10^{18}) \text{ J} $
$ \Delta E = (6.024 \times 10^{-38}) \times (25 \times 10^{18} - 0.25 \times 10^{18}) \text{ J} $
$ \Delta E = (6.024 \times 10^{-38}) \times (24.75 \times 10^{18}) \text{ J} $
$ \Delta E \approx 1.491 \times 10^{-18} \text{ J} $
-
Convert the energy difference per electron (in Joules) to kilojoules per mole (kJ/mol).
The conversion uses Avogadro's number (NA) and the factor 1000 J/kJ:
$ \Delta E (\text{kJ/mol}) = \frac{\Delta E (\text{J}) \times N_A}{1000} $
$ \Delta E (\text{kJ/mol}) = \frac{(1.491 \times 10^{-18} \text{ J}) \times (6.022 \times 10^{23} \text{ mol}^{-1})}{1000 \text{ J/kJ}} $
$ \Delta E (\text{kJ/mol}) \approx \frac{8.978 \times 10^5 \text{ J/mol}}{1000 \text{ J/kJ}} $
$ \Delta E (\text{kJ/mol}) \approx 897.8 \text{ kJ/mol} $
Conclusion
The calculated energy difference is approximately 897.8 kJ/mol. This value lies between 896 and 900 kJ/mol, consistent with the provided answer range.
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 ________
