Chapter 11, Problem 11A.1ST

Interpretation Introduction

Interpretation:

The absorption of the solution and molar absorption coefficient of tyrosine at $240\text{nm}$ has to be calculated.  The transmittance through cell of length $1.0\text{mm}$ has to be calculated.

Concept introduction:

Spectroscopy method is used to identify the structure of the molecule.  It is based on the interactions between matter and electromagnetic radiations.  An electronic state of energy has its own vibrational states.  The energy between the electronic states is large followed by vibrational states and then rotational states.  During an electronic transition, electron from ground state moves straight to the excited state keeping the internuclear distance constant.

Answer to Problem 11A.1ST

The absorption of the solution and molar absorption coefficient of tyrosine at $240\text{nm}$ is $\underset{\_}{0.85}$ and $\underset{\_}{1.7\times 10^{4}mo{l}^{-1}d{m}^{3}c{m}^{-1}}$, respectively.  The transmittance through cell of length $1.0\text{mm}$ is $\underset{\_}{0.67}$.

Explanation of Solution

The relationship between transmittance and absorption is given as shown below.

    $A=-\log T$        (1)

Where,

• $A$ is the absorption.
• $T$ is the transmittance.

The transmittance of solution containing tyrosine is $0.14$.

Substitute the value of transmittance in equation (1).

    $\begin{array}{c}A=-\log \left(0.14\right)\\ =\underset{\_}{0.85}\end{array}$

The molar absorption coefficient is calculated by the formula shown below.

    $\epsilon =\frac{A}{\left[\text{J}\right]L}$        (2)

Where,

• $\left[\text{J}\right]$ is the concentration of tyrosine in solution.
• $L$ is the thickness of cell.

The concentration of tyrosine in solution $0.10\text{mmol}{\text{dm}}^{-3}$.

The thickness of cell is $5.0\text{mm}$.

Convert $5.0\text{mm}$ to $\text{cm}$ as shown below.

  $\begin{array}{c}5.0\text{mm}=\frac{5.0}{10}\text{cm}\\ =0.5\text{cm}\end{array}$

Substitute the value of absorption, thickness of cell and concentration of tyrosine in solution in equation (2).

    $\begin{array}{c}\epsilon =\frac{0.85}{0.10\text{mmol}{\text{dm}}^{-3}\times \frac{1\text{mol}}{1000\text{mmol}}\times 0.5\text{cm}}\\ =\frac{0.85}{0.50\times {10}^{-4}\text{mol}{\text{dm}}^{-3}\text{cm}}\\ =\underset{\_}{1.7\times 10^{4}mo{l}^{-1}d{m}^{3}c{m}^{-1}}\end{array}$

Rearrange equation (2) to calculate the value of absorption.

    $A=\epsilon \left[\text{J}\right]L$        (3)

Substitute the value of absorption, thickness of cell $\left(1\text{mm}\right)$ and concentration of tyrosine in solution in equation (3).

    $\begin{array}{c}A=1.7\times {10}^4{\text{mol}}^{-1}{\text{dm}}^3{\text{cm}}^{-1}\times 0.10\text{mmol}{\text{dm}}^{-3}\times \frac{1\text{mol}}{1000\text{mmol}}\times 1\text{mm}\times \frac{1\text{cm}}{10\text{mm}}\\ =1.7\times {10}^4{\text{mol}}^{-1}{\text{dm}}^3{\text{cm}}^{-1}\times 0.10\times {10}^{-3}\text{mol}{\text{dm}}^{-3}\times 0.1\text{cm}\\ =0.17\end{array}$

Rearrange equation (1) to calculate the value of transmittance.

    $\begin{array}{c}T={10}^{-A}\\ =\underset{\_}{0.67}\end{array}$