Quantum Chemistry/Example 5

For a particle (assume the particle is an electron with 1 quantum number.) in a 1D box of length 5 cm, the equation of energy levels of a particle in a 1D box is,

$$E_n = \frac{h^2n^2}{8mL^2} $$

a. If the length of the 1D box is increased to 10 cm, what is the change in the energy level of this particle in the box?

b. If the length of the 1D box is decreased to 2 cm, what is the change in the energy level of this particle in the box?

c. Explain the effect of length changes on the energy levels of particles in a 1D box.

Solutions
The energy levels of a particle in 1D box:

$$E_n = \frac{h^2n^2}{8mL^2} $$

where h is Planck's constant equal to $$6.626 \times 10^{-34}\text{ J s}$$

Because this particle is an electron with 1 quantum number:

m is the mass of this particle equal to $$9.109 \times 10^{-31}\text{ kg}$$

n is the quantum number of the particle equal to 1

L is the length of this 1D box equal to $$5 \text{ cm}$$

for this question,

$$\Delta{E} = E_{nf} - E_{ni}$$

$$= \left[\frac{h^2n^2} {8m} \frac{1}{L_f^2}\right] - \left[\frac{h^2n^2} {8m} \frac{1}{L_i^2}\right]$$

$$= \left[\frac{h^2n^2} {8m}\right] \times \left[\frac{1} {L_f^2} - \frac{1}{L_i^2}\right]$$

a. for part a, the initial length of this 1D box: $$L_i = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$$, the final length of this 1D box: $$L_f = 10 \text{ cm} = 10 \times 10^{-2} \text{ m}$$

$$\Delta{E} = \left[\frac{h^2n^2} {8m}\right] \times \left[\frac{1} {L_f^2} - \frac{1}{L_i^2}\right]$$

$$= \left[\frac{(6.626 \times 10^{-34}\text{ J s})^2 (1)^2 } {(8)(9.109 \times 10^{-31}\text{ kg})}\right] \left[\frac{1} {(10 \times 10^{-2} \text{m})^2} - \frac{1} {(5 \times 10^{-2} \text{m})^2} \right]$$

$$= -1.807 \times 10^{-35} \text{ J}$$

b. for part b, the initial length of this 1D box: $$L_i = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$$, the final length of this 1D box: $$L_f = 2 \text{ cm} = 2 \times 10^{-2} \text{ m}$$

$$\Delta{E} = \left[\frac{h^2n^2} {8m}\right] \times \left[\frac{1} {L_f^2} - \frac{1}{L_i^2}\right]$$

$$= \left[\frac{(6.626 \times 10^{-34}\text{ J s})^2 (1)^2 } {(8)(9.109 \times 10^{-31}\text{ kg})}\right] \left[\frac{1} {(2 \times 10^{-2} \text{m})^2} - \frac{1} {(5 \times 10^{-2} \text{m})^2} \right]$$

$$= 1.265 \times 10^{-34} \text{ J}$$

c. Based on the calculations in part a and b, the energy level of the particles in the 1D box is decreased by the increase of this 1D box length, and the energy level of the particles in the 1D box is increased by the decrease of this 1D box length. Therefore, the energy level of the particles in the 1D box is negatively related to the length of this 1D box.