Chapter 14, Problem 14D.2P

(a)

Interpretation Introduction

Interpretation:

The expression for the root-mean-square separation has to be derived.  The root-mean-square separation for a flexible chain with $N=4000$ and $l=154\text{pm}$ has to be stated.

Concept introduction:

The one dimensional freely jointed chain is said to be the primary structure of the macromolecule.  The primary structure of the macromolecule is the sequence of the residues which make up the macromolecule.

Answer to Problem 14D.2P

The expression for the root-mean-square separation has been derived.  The root-mean-square separation for a flexible chain with $N=4000$ and $l=154\text{pm}$ is $\underset{\_}{9739.8pm}$.

Explanation of Solution

The probability function for a three dimension flexible chain with radius $r$ is given by the equation as shown below.

  $f\left(r\right)=4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3{r}^2{\text{e}}^{-a^2{r}^2}$        (1)

In the above equation the constant $a$ is given by ${\left(\frac{3}{2N{l}^2}\right)}^{1/2}$.

The square of the root mean square separation is given by the equation as shown below.

  $R_{rms}{}^2=\langle r^2\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^{2}f\left(r\right)dr}$        (2)

Substitute equation (1) in equation (2) and integrate as shown below.

  $\begin{array}{c}{R}_{rms}{}^2={\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^{2}f\left(r\right)dr}\\ =4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^4{\text{e}}^{-a^2{r}^2}}\end{array}$

Use the standard integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^4{\text{e}}^{-k{x}^2}}=\frac{3}{8k^2}{\left(\frac{\text{π}}{k}\right)}^{1/2}$ in the above equation.

  $\begin{array}{c}{R}_{\text{rms}}{}^2=4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3\times \frac{3}{8a^4}{\left(\frac{\text{π}}{a^2}\right)}^{1/2}\\ =\frac{3{\text{π}}^{1/2}}{2a^2}\end{array}$

Substitute the value of $a$ in the above equation.

  $\begin{array}{c}{R}_{\text{rms}}{}^2=\frac{3{\text{π}}^{1/2}}{2}\left(\frac{2N{l}^2}{3}\right)\\ R_{\text{rms}}{}^2=N{l}^2\\ R_{\text{rms}}=N^{1/2}l\end{array}$

For the value of root-mean-square separation for a flexible chain, substitute with $N=4000$ and $l=154\text{pm}$ in the above equation.

  $\begin{array}{c}{R}_{\text{rms}}={\left(4000\right)}^{1/2}154\text{pm}\\ =\underset{\_}{9739.8pm}\end{array}$

Therefore, the root-mean-square separation for a flexible chain with $N=4000$ and $l=154\text{pm}$ is $\underset{\_}{9739.8pm}$.

(b)

Interpretation Introduction

Interpretation:

The expression for the mean separation of the ends has to be derived.  The mean separation of the ends for a flexible chain with $N=4000$ and $l=154\text{pm}$ has to be stated.

Concept introduction:

The same concept introduction as in subpart (a).

Answer to Problem 14D.2P

The expression for the mean separation of the ends has been derived.  The mean separation of the ends for a flexible chain with $N=4000$ and $l=154\text{pm}$ is $\underset{\_}{8975.73pm}$.

Explanation of Solution

The probability function for a three dimension flexible chain with radius $r$ is given by the equation as shown below.

  $f\left(r\right)=4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3{r}^2{\text{e}}^{-a^2{r}^2}$        (1)

In the above equation the constant $a$ is given by ${\left(\frac{3}{2N{l}^2}\right)}^{1/2}$.

The mean separation of the ends is given by the equation as shown below.

  $R_{\text{mean}}=\langle r\rangle ={\displaystyle \underset{0}{\overset{\infty }{\int }}rf\left(r\right)dr}$        (3)

Substitute equation (1) in equation (3) and integrate as shown below.

  $\begin{array}{c}{R}_{mean}={\displaystyle \underset{0}{\overset{\infty }{\int }}rf\left(r\right)dr}\\ =4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3{\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^3{\text{e}}^{-a^2{r}^2}}\end{array}$

Use the standard integral ${\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^3{\text{e}}^{-k{x}^2}}=\frac{1}{2k^2}$ in the above equation.

  $\begin{array}{c}{R}_{\text{mean}}=4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3\times \frac{1}{2a^4}\\ =\frac{2}{{\text{π}}^{1/2}a}\end{array}$

Substitute the value of $a$ in the above equation.

  $\begin{array}{c}{R}_{\text{mean}}=\frac{2}{{\text{π}}^{1/2}}{\left(\frac{2N{l}^2}{3}\right)}^{1/2}\\ R_{\text{mean}}={\left(\frac{8N}{3\text{π}}\right)}^{1/2}l\end{array}$

For the value of mean separation of the ends for a flexible chain, substitute with $N=4000$ and $l=154\text{pm}$ in the above equation.

  $\begin{array}{c}{R}_{\text{mean}}={\left(\frac{8\times 4000}{3\times \text{3}\text{.14}}\right)}^{1/2}\times 154\text{pm}\\ =58.28\times 154\text{pm}\\ =\underset{\_}{8975.73pm}\end{array}$

Therefore, the mean separation of the ends for a flexible chain with $N=4000$ and $l=154\text{pm}$ is $\underset{\_}{8975.73pm}$.

(c)

Interpretation Introduction

Interpretation:

The most probable separation has to be derived.  The most probable separation for a flexible chain with $N=4000$ and $l=154\text{pm}$ has to be stated.

Concept introduction:

The same concept introduction as in subpart (i).

Answer to Problem 14D.2P

The most probable separation is derived.  The most probable separation for a flexible chain with $N=4000$ and $l=154\text{pm}$ is $\underset{\_}{7952.52pm}$.

Explanation of Solution

The most probable separation is given by the value of $r\to R^{*}$ at which $f\left(r\right)$ is maximum.

The value at which $f\left(r\right)$ is maximum is as shown below.

  $\frac{df}{dr}=0$        (4)

Substitute equation (1) in equation (4).

  $\begin{array}{c}\frac{df}{dr}=\frac{d}{dr}\left(4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3{r}^2{\text{e}}^{-a^2{r}^2}\right)\\ =4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3\left(2r{\text{e}}^{-a^2{r}^2}-2r^3{a}^2{\text{e}}^{-a^2{r}^2}\right)\\ =4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3\left({\text{e}}^{-a^2{r}^2}\right)\left(2r-2r^3{a}^2\right)\end{array}$

Equate the above expression with zero and solve for $r\to R^{*}$ as shown below.

  $\begin{array}{c}4\text{π}{\left(\frac{a}{{\text{π}}^{1/2}}\right)}^3\left({\text{e}}^{-a^2{\left(R^{*}\right)}^2}\right)\left(2R^{*}-2{\left(R^{*}\right)}^3{a}^2\right)=0\\ \left(2R^{*}-2{\left(R^{*}\right)}^3{a}^2\right)=0\\ R^{*}=\frac{1}{a}\end{array}$

Substitute the value of $a$ in the above equation.

  $\begin{array}{c}{R}^{*}={\left(\frac{2N{l}^2}{3}\right)}^{1/2}\\ ={\left(\frac{2N}{3}\right)}^{1/2}l\end{array}$

For the value of most probable separation for a flexible chain, substitute with $N=4000$ and $l=154\text{pm}$ in the above equation.

  $\begin{array}{c}{R}^{*}={\left(\frac{2\times 4000}{3}\right)}^{1/2}\times 154\text{pm}\\ =51.63\times 154\text{pm}\\ =\underset{\_}{7952.52pm}\end{array}$

Therefore, the most probable separation for a flexible chain, substitute with $N=4000$ and $l=154\text{pm}$ is $\underset{\_}{7952.52pm}$.