# Physical Chemistry Questions and Answers – Particle in a Three-Dimension Box

This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Particle in a Three-Dimension Box”.

1. What is the classical Hamiltonian expression for a particle moving in 3-dimensions with no potential energy?
a) H = 0
b) H = $$\frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m}$$
c) H = $$\frac{-\hbar^2}{2m}$$ ∇2
d) H = $$\frac{-\hbar^2}{2m}$$ ∇2 + V(x,y,z)

Explanation: H = $$\frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m}$$ is the classical Hamiltonian expression if kinetic energy is replaced by momentum in all 3 directions. H = $$\frac{-\hbar^2}{2m}$$ ∇2 is the quantum mechanical expression for the same and there is no potential energy term.

2. What is the quantum mechanical Hamiltonian expression for a particle moving in 3-dimesnions with potential energy V?
a) H = 0
b) H = $$\frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{p_z^2}{2m}$$
c) H = $$\frac{-\hbar^2}{2m}$$ ∇2
d) H = $$\frac{-\hbar^2}{2m}$$ ∇2 + V(x,y,z)

Explanation: H = $$\frac{-\hbar^2}{2m}$$ ∇2 + V(x,y,z) is the quantum mechanical Hamiltonian for a particle moving with kinetic and potential energy. This is shown with the Laplacian operator (differential of position in all three directions). There is no separate operator for potential energy in quantum mechanics.

3. The wavefunction for a 3-D particle in a box is given by φ = A sin⁡ $$\frac{n_x \pi x}{a}$$ sin⁡ $$\frac{n_y \pi x}{a}$$ sin⁡ $$\frac{n_z \pi x}{a}$$. What is the value of A for normalization?
a) $$(\frac{8}{abc})^{\frac{1}{2}}$$
b) 1
c) $$(\frac{3\pi^2}{abc})^{\frac{1}{2}}$$
d) $$(\frac{abc}{8\pi})^{\frac{1}{2}}$$

Explanation: $$\int_{-\infty}^\infty$$ $$\int_{-\infty}^\infty$$ $$\int_{-\infty}^\infty$$ A sin⁡ $$\frac{n_x \pi x}{a}$$ sin⁡ $$\frac{n_y \pi x}{a}$$ sin⁡ $$\frac{n_z \pi x}{a}$$ dxdydz. Integrating each term, equating to one, and removing the coefficients gives A = $$(\frac{8}{abc})^{\frac{1}{2}}$$

4. After substituting the eigenfunctions of φ in a 3-D box, what is the energy obtained for a particle in a 3-D box?
a) E = $$\frac{h^2 π^2 n^2}{8ma}$$
b) E = $$\frac{h^2}{8m}(\frac{n_x^2}{a})$$
c) E = $$\frac{h^2 π^2}{8mabc}$$
d) E = $$\frac{h^2}{8m}(\frac{n_x^2}{a}+\frac{n_y^2}{b}+\frac{n_z^2}{c})$$

Explanation: Replacing φ = $$(\frac{8}{abc})^{\frac{1}{2}}$$ sin⁡ $$\frac{n_y \pi x}{a}$$ sin⁡ $$\frac{n_z \pi x}{a}$$in the Schrodinger equation
$$\frac{-\hbar^2}{2m}$$ ∇2 = Eφ gives E = $$\frac{h^2}{8m}(\frac{n_x^2}{a}+\frac{n_y^2}{b}+\frac{n_z^2}{c})$$

5. What is the degeneracy temperature for a carbon atom in a three-dimensional cubic box? (Assume the gas molecules to behave like point particles with no intermolecular forces)
a) T = $$\frac{h^2(n^2_x+n_y^2+n_z^2)}{12ma^2k}$$
b) T = $$\frac{h^2}{8m}(\frac{n_x^2}{a}+\frac{n_y^2}{b}+\frac{n_z^2}{c})$$
c) T = $$\frac{h^2}{8m}(\frac{n_x^2}{a}+\frac{n_y^2}{b}+\frac{n_z^2}{c})$$
d) T = $$\frac{3h^2}{16km} (\frac{n_x^2}{a}+\frac{n_y^2}{b}+\frac{n_z^2}{c})$$

Explanation: The degeneracy energy in a quantum state E = $$\frac{h^2}{8m}(\frac{n_x^2}{a}+\frac{n_y^2}{b}+\frac{n_z^2}{c})$$ is equated to thermal energy E = $$\frac{3}{2}$$kT. Since this is a cubic box where a = b = c, $$\frac{3}{2}$$kT = $$\frac{h^2}{8m}(\frac{n_x^2}{a}+\frac{n_y^2}{b}+\frac{n_z^2}{c})$$. Solving for T gives T = $$\frac{h^2(n^2_x+n_y^2+n_z^2)}{12ma^2k}$$.

Sanfoundry Global Education & Learning Series – Physical Chemistry.

To practice all areas of Physical Chemistry, here is complete set of Multiple Choice Questions and Answers.