Question

A given polymer, whose mean square radius of gyration is 7.54·10^3 Å^2, has at its ends two reactive groups capable of emitting light when they are less than 5 Å away (regardless of which one it is in particular). this distance). The emission is proportional to the number of molecules that at each moment meet the aforementioned condition and we consider ideal dilute solutions, or “theta” in which the probability of the two terminal groups of different molecules approaching is small. Obtain the ratio between the amount emitted by a solution of a practically monodisperse sample of said polymer and another similar one, and of the same concentration by weight, but in which the mean square radius of gyration of the molecules is 1.32·10^ 4 Å^2.

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Answer to a math question A given polymer, whose mean square radius of gyration is 7.54·10^3 Å^2, has at its ends two reactive groups capable of emitting light when they are less than 5 Å away (regardless of which one it is in particular). this distance). The emission is proportional to the number of molecules that at each moment meet the aforementioned condition and we consider ideal dilute solutions, or “theta” in which the probability of the two terminal groups of different molecules approaching is small. Obtain the ratio between the amount emitted by a solution of a practically monodisperse sample of said polymer and another similar one, and of the same concentration by weight, but in which the mean square radius of gyration of the molecules is 1.32·10^ 4 Å^2.

$Expert avatar$

Cristian

4.7

119 Answers

To calculate the ratio

of intensities between the two polymers with different radii of gyration, we need to evaluate the integrals involving the probabilities

P_1(r)

and

P_2(r)

over the range from 0 to 5 Å.

Given:

R_{g1}^2 = 7.54 \times 10^3 \, Å^2

R_{g2}^2 = 1.32 \times 10^4 \, Å^2

We know that

R_e^2 = 6R_g^2

. Therefore, the mean square end-to-end distances for the two polymers are:

R_{e1}^2 = 6 \times 7.54 \times 10^3 = 4.524 \times 10^4 \, Å^2

R_{e2}^2 = 6 \times 1.32 \times 10^4 = 7.92 \times 10^4 \, Å^2

The probability density functions

P_1(r)

and

P_2(r)

are given by:

P_1(r) = \left(\frac{3}{2\pi \times 4.524 \times 10^4}\right)^{3/2} \exp\left(-\frac{3r^2}{2 \times 4.524 \times 10^4}\right)

P_2(r) = \left(\frac{3}{2\pi \times 7.92 \times 10^4}\right)^{3/2} \exp\left(-\frac{3r^2}{2 \times 7.92 \times 10^4}\right)

Now, we need to calculate:

R = \frac{I_1}{I_2} = \frac{\int_{0}^{5} P_1(r) \, 4\pi r^2 dr}{\int_{0}^{5} P_2(r) \, 4\pi r^2 dr}

Integrating, we get:

R = \frac{\int_{0}^{5} P_1(r) \, 4\pi r^2 dr}{\int_{0}^{5} P_2(r) \, 4\pi r^2 dr} = 2.32

\boxed{R = 2.32}

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