Question

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At what value of n does the energy of the molecule reach 1 2 kT at 300K, and what is the separation of this level from the one immediately below

Answer 1

Answer:

The separation between the two lowest levels =  $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$J

Explanation:

Knowing the formula

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

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 =  $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

by the end is necessary to calculate the separation of the level

En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

              = 1.8 * $10^{-30}$J

Answer 2

a. The energy difference between the lowest energy level (n = 1 and n = 2) is $1.25\; \times \;10^{-39}\;Joules$

b. The n value for which the energy reaches $\frac{1}{2} kT$ is $2.24 \times 10^9$

c. The separation between the lowest levels from the one immediately below (n and n - 1) is $1.85\times 10^{-30}\;Joules$

Given the following data:

• Length = 5.0 cm to m = 0.05 meter.

• Temperature = 300 K

Scientific data:

• Planck constant = $6.626 \times 10^{-34}\;J.s$

• Boltzmann constant (k) = $1.38 \times 10^{-23}\;J/K$

• Atomic mass of $O_2$ = 32 amu

First of all, we would determine the mass of $O_2$ molecule:

$Mass \;of\;O_2 = 32 \times 1.6605 \times 10^{-27}\\\\Mass \;of\;O_2 =5.3 \times 10^{-26}\;kg$

For a one-dimensional particle in a container, the particle has only kinetic energy (V = 0) and the permitted energies are given by the formula:

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

Where:

• m is the mass of the particle.

• L is the length of container.

• h is Planck constant.

• n is the energy level.

a. The energy difference between the lowest energy level (n = 1 and n = 2) is:

$\Delta E = E_2 - E_1 = \frac{2^2h^2}{8mL^2}-\frac{1^2h^2}{8mL^2}\\\\\Delta E = \frac{3^2h^2}{8mL^2}$

Substituting the given parameters into the formula, we have;

$\Delta E = \frac{3\; \times \;(6.626\; \times \;10^{-34})^2}{8 \;\times \;5.3 \;\times \;10^{-26}\; \times \;0.05^2} \\\\\Delta E = \frac{3\; \times \;4.39\; \times \;10^{-67}}{4.24 \;\times \;10^{-25}\; \times \;0.0025} \\\\\Delta E = \frac{1.32\; \times \;10^{-66}}{1.06 \;\times \;10^{-27}} \\\\\Delta E =1.25\; \times \;10^{-39}\;Joules$

Note: The lowest-energy state (zero-point energy) is given by the formula:

$\frac{h^2}{8mL^2} =4.14\times 10^{-40}\;Joules$

b. The n value for which the energy reaches $\frac{1}{2} kT$ is given by the formula:

$E_n = \frac{n^2h^2}{8mL^2} = \frac{1}{2} kT\\\\n^2 (4.14\times 10^{-40})=\frac{1}{2} \times 1.38 \times 10^{-23} \times 300\\\\n^2 (4.14\times 10^{-40})=2.07 \times 10^{-21} \\\\n^2 =\frac{2.07 \times 10^{-21} }{4.14\times 10^{-40}} \\\\n^2=5 \times 10^{18}\\\\n=\sqrt{5 \times 10^{18}}\\\\n=2.24 \times 10^9$

Lastly, we would calculate the separation between the lowest levels from the one immediately below (n and n - 1):

$E_n-E_{n-1} =(n^2 -[n-1]^2 )\frac{h^2}{8mL^2} \\\\E_n-E_{n-1} =(n^2 - [n^2-n -n+1])\frac{h^2}{8mL^2}\\\\E_n-E_{n-1}=(2n-1)\frac{h^2}{8mL^2}\\\\E_n - E_{n-1}=(2n-1)\frac{h^2}{8mL^2}\\\\E_n - E_{n-1}=(2[2.24 \times 10^9]-1) \times 4.14 \times 10^{-40}\\\\E_n - E_{n-1}=(4.48 \times 10^{12}-1) \times 4.14 \times 10^{-40}\\\\E_n - E_{n-1}=(4,479,999,999,999) \times 4.14 \times 10^{-40}\\\\E_n - E_{n-1}=1.85\times 10^{-30}\;J$

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