Macromolecule Chain End-to-End Distance Probability Calculation

This problem asks for the probability of finding the ends of a macromolecule chain within a specific distance range. The macromolecule chain is modeled as a Gaussian chain in three dimensions. The probability density function for the end-to-end distance magnitude \(r\) of a Gaussian chain is given by:

\[P(r) = 4\pi r^2 \left(\frac{3}{2\pi N l^2}\right)^{3/2} e^{-\frac{3r^2}{2N l^2}}\]

Where:

• \(N\) is the number of segments.
• \(l\) is the length of each segment.
• \(r\) is the end-to-end distance.

We are given the following values:

• Number of segments, \(N = 100\)
• Segment length, \(l = 150 \, \text{pm}\). Let's convert this to meters: \(l = 150 \times 10^{-12} \, \text{m}\).
• Distance, \(r = 3.00 \, \text{nm}\). Let's convert this to meters: \(r = 3.00 \times 10^{-9} \, \text{m}\).
• Thickness of the shell, \(\Delta r = 10.0 \, \text{pm}\). Let's convert this to meters: \(\Delta r = 10.0 \times 10^{-12} \, \text{m}\).

The probability of finding the ends within the distance range \(r\) to \(r + \Delta r\) is approximately \(P(r) \Delta r\), assuming \(\Delta r\) is small.

First, let's calculate the term \(Nl^2\):

\(Nl^2 = 100 \times (150 \times 10^{-12} \, \text{m})^2 = 100 \times (22500 \times 10^{-24} \, \text{m}^2) = 2.25 \times 10^{-18} \, \text{m}^2\)

Next, calculate the exponent term argument, \(\frac{3r^2}{2N l^2}\):

\(r^2 = (3.00 \times 10^{-9} \, \text{m})^2 = 9.00 \times 10^{-18} \, \text{m}^2\)

\(\frac{3r^2}{2N l^2} = \frac{3 \times 9.00 \times 10^{-18} \, \text{m}^2}{2 \times 2.25 \times 10^{-18} \, \text{m}^2} = \frac{27.0 \times 10^{-18}}{4.5 \times 10^{-18}} = 6\)

So the exponential term is \(e^{-6}\).

Now, calculate the constant prefactor \(\left(\frac{3}{2\pi N l^2}\right)^{3/2}\). Let \(b = \frac{3}{2N l^2}\):

\(b = \frac{3}{2 \times 2.25 \times 10^{-18} \, \text{m}^2} = \frac{3}{4.5 \times 10^{-18}} \, \text{m}^{-2} = \frac{2}{3} \times 10^{18} \, \text{m}^{-2}\)

The prefactor is \(\left(\frac{b}{\pi}\right)^{3/2}\):

\(\left(\frac{b}{\pi}\right)^{3/2} = \left(\frac{\frac{2}{3} \times 10^{18} \, \text{m}^{-2}}{\pi}\right)^{3/2} = \left(\frac{2}{3\pi} \times 10^{18}\right)^{3/2} \, \text{m}^{-3}\)

Using \(\pi \approx 3.14159\), \(\frac{2}{3\pi} \approx \frac{2}{9.42477} \approx 0.212207\). So, \(\left(\frac{b}{\pi}\right)^{3/2} \approx (0.212207 \times 10^{18})^{1.5} \, \text{m}^{-3} = (0.212207)^{1.5} \times (10^{18})^{1.5} \, \text{m}^{-3} \approx 0.097753 \times 10^{27} \, \text{m}^{-3}\).

Now, calculate the probability density \(P(r)\) at \(r = 3.00 \, \text{nm}\):

\(P(r) = 4\pi r^2 \left(\frac{b}{\pi}\right)^{3/2} e^{-br^2}\)

\(P(r) = 4\pi (9.00 \times 10^{-18} \, \text{m}^2) (0.097753 \times 10^{27} \, \text{m}^{-3}) e^{-6}\)

Using \(e^{-6} \approx 0.00247875\):

\(P(r) \approx 4\pi \times 9 \times 10^{-18} \times 0.097753 \times 10^{27} \times 0.00247875 \, \text{m}^{-1}\)

\(P(r) \approx (36\pi \times 0.097753 \times 0.00247875) \times 10^{(-18+27)} \, \text{m}^{-1}\)

\(P(r) \approx (113.097 \times 0.097753 \times 0.00247875) \times 10^9 \, \text{m}^{-1}\)

\(P(r) \approx (11.0569 \times 0.00247875) \times 10^9 \, \text{m}^{-1}\)

\(P(r) \approx 0.027407 \times 10^9 \, \text{m}^{-1} = 2.7407 \times 10^7 \, \text{m}^{-1}\)

Finally, calculate the probability \(\Delta P = P(r) \Delta r\):

\(\Delta P \approx (2.7407 \times 10^7 \, \text{m}^{-1}) \times (10.0 \times 10^{-12} \, \text{m})\)

\(\Delta P \approx 2.7407 \times 10^{(7 + (-12))} = 2.7407 \times 10^{-5}\)

Let's recheck the multiplication by \(\Delta r\):

\(\Delta P \approx (2.7407 \times 10^7) \times (10 \times 10^{-12}) = 2.7407 \times 10^{7+1-12} = 2.7407 \times 10^{-4}\)

The calculated probability using the standard Gaussian chain model is approximately \(2.74 \times 10^{-4}\).

Reviewing the provided options:

1. \(1.92 \times 10^{-4}\)
2. \(1.92 \times 10^{-3}\)
3. \(1.92\)
4. \(1.92 \times 10^{-5}\)

The calculated value \(2.74 \times 10^{-4}\) is closest in magnitude to option 1 \(1.92 \times 10^{-4}\). However, the provided correct answer corresponds to Option 2.

If we assume the intended probability is \(1.92 \times 10^{-3}\), then the corresponding probability density at \(r=3.00 \, \text{nm}\) would need to be \(\frac{1.92 \times 10^{-3}}{10 \times 10^{-12} \, \text{m}} = 1.92 \times 10^8 \, \text{m}^{-1}\). This value is higher than what the standard Gaussian chain formula predicts at this distance.

Based on the standard formula and given parameters, the calculated probability is \(2.74 \times 10^{-4}\).