Question

Buckminsterfullerene, C₆₀, is a large molecule consisting of 60 carbon atoms connected to form a hollow sphere. The diameter of a C₆₀ molecule is about 7×10^-10 m. It has been hypothesized that C₆₀ molecules might be found in clouds of interstellar dust, which often contain interesting chemical compounds. The temperature of an interstellar dust cloud may be very low; around 3 K. Suppose you are planning to try to detect the presence of C60 in such a cold dust cloud by detecting photons emitted when molecules undergo transitions from one rotational energy state to another. Approximately, what is the highest-numbered rotational level from which you would expect to observe emissions? Rotational levels are l= 0, 1, 2, 3, …

Short answer

The rotational level for the molecule at the given temperature is l = 9 .

Step-by-step solution

Identification of given data

• The diameter of the C₆₀ molecule is,$d=7x{10}^{-10}m$ .
• The radius of the C₆₀ molecule is,$r=3.5\times {10}^{-10}m$.
• The given temperature is, T = 3K .

Concept of Wein’s displacement law

Black-body radiation peaks at various wavelengths depending on temperature, according to Wien's displacement equation.

Wein’s displacement is given by,

$\lambda =\frac{b}{T}$

Here, b is the constant of proportionality and is the temperature.

Determination of rotational energy

The wavelength of the wave can be evaluated using equation (i); we get,

$$\begin{gathered} \lambda =\frac{2.898\times {10}^{-3}m.K}{3K} \\ \lambda =9.66\times {10}^{-4}m \end{gathered}$$

The energy of the photon is given by,

$E=\frac{hc}{\lambda }$

Here, h is Planck’s constant, c is the speed of light, $\lambda$is the wavelength of the light.

Substitute values in the above equation, and we get,

$$\begin{gathered} E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m/s\right)}{9.66\times {10}^{-4}m} \\ E=2.057\times {10}^{-22}J \end{gathered}$$

The moment of inertia is evaluated using the equation,

$I=\left(\frac{2}{3}M{r}^2\right)$

Here $M=60\times 12\times 1.67\times {10}^{-27}Kg$ is the mass of the C₆₀ atom.

Substitute the value in the above equation, and we get,

$\begin{array}{ll}I=\left(\frac{2}{3}\right)(60\times 12\times 167\times {10}^{-27}& \mathrm{kg}){(3.5\times {10}^{-10}m)}^2\\ l=982\times {10}^{-44}Kg{m}^2& \end{array}$

The rotational energy can be evaluated using the equation,

$E=\frac{h^2}{2l}l(1+I)$

Here l is the value of the rotational level.

Substitute value and re-arranging the equation,

$$\begin{gathered} 2.057\times {10}^{-22}J=\frac{{\left(6.67\times {10}^{-34}J.s\right)}^2}{2\times 9.82\times {10}^{-44}Kg.m^2}l(1+l) \\ {l}^2+l-92=0 \\ l\approx 9 \end{gathered}$$

Thus, the highest rotational level for the molecule at a given temperature is 9 .