One fairly crude method of determining the size of a molecule is to treat the molecule as an infinite square well (a box) with an electron trapped inside,and to measure the wavelengths of emitted photons.If the photon emitted during the n = 2 to n = 1 transition has wavelength 1940 nm,what is the width of the molecule? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34J ∙ s,
M_el = 9.11 × 10^-31 kg)

Question

Quizplus · Accepted Answer

The energy difference between the n = 2 and n = 1 levels in an infinite square well is given by $\Delta E = E_2 - E_1 = \frac{3h^2}{8mL^2}$, where $L$ is the width of the well (molecule in this case), $m$ is the mass of the electron, and $h$ is Planck's constant. The energy of the emitted photon, which corresponds to this energy difference, can also be expressed as $E = \frac{hc}{\lambda}$, where $\lambda$ is the wavelength of the photon, $c$ is the speed of light, and $h$ is Planck's constant.Given $\lambda = 1940 \, \text{nm} = 1940 \times 10^{-9} \, \text{m}$, $c = 3.00 \times 10^8 \, \text{m/s}$, $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$, and $m = 9.11 \times 10^{-31} \, \text{kg}$, we can find $L$ by equating the two expressions for energy and solving for $L$.First, find the energy of the photon: $E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^8}{1940 \times 10^{-9}} \, \text{J}$.Then, equate this to the expression for $\Delta E$ and solve for $L$: $L = \sqrt{\frac{3h^2}{8mE}}$.Substituting the values and solving for $L$ gives a width close to 1.33 nm, which corresponds to choice A.

One Fairly Crude Method of Determining the Size of a Molecule