Question

Benzene, C6H6, has a cyclic structure where the carbon atoms make a hexagon. The π electrons in the cyclic molecule can be approximated as having two-dimensional rotational motion. Calculate the diameter of this “electron ring” if it is assumed that a transition occurring at 260.0 nm corresponds to an electron going from m = 3 to m = 4.

Answer 1

To calculate the diameter of the "electron ring" in benzene, we can use the formula for the diameter of a circular path:

d = λ / (2π)

where: d is the diameter of the circular path, λ is the wavelength of the transition, and π is a mathematical constant approximately equal to 3.14159.

In this case, the transition wavelength (λ) is given as 260.0 nm.

Substituting the values into the formula, we get:

d = 260.0 nm / (2π)

To convert the wavelength from nanometers (nm) to meters (m), we divide by 10^9:

d = (260.0 × 10^(-9) m) / (2π)

Now we can calculate the diameter:

d ≈ 4.142 × 10^(-8) m / π

Using the approximation of π as 3.14159, we can evaluate the expression:

d ≈ 1.317 × 10^(-8) m

Therefore, the diameter of the "electron ring" in benzene, based on the given transition wavelength, is approximately 1.317 × 10^(-8) meters.