Chapter 30, Problem 30.12QAP

Interpretation Introduction

(a)

Interpretation:

The equation for the calibration curve, standard deviations of the slope, intercept from the data of doxorubicin by LIF, its peak value and the R² value needs to be determined.

Concept introduction:

The standard deviation is to tell how the measurements of group are spread out from average. In other words, it defines how far a set of numbers lie apart. If the value obtained is low, it means that most of the numbers are close to average, but if the value obtained is high, it means the number are spread out.

Answer to Problem 30.12QAP

The standard deviation of m is 0.015813, standard deviation of intercept is 0.374481. The value of R² is 0.99821.

Explanation of Solution

The graphical representation is as below:

From the equation of straight line,

Slope m = 0.835066

Intercept, b = 0.026972

Standard deviation of m= 0.015813

Standard deviation of b = 0.374481

R² = 0.99821

Interpretation Introduction

(b)

Interpretation:

The equation should be rearranged, and concentration should be expressed in terms of measured area.

Concept introduction:

For a straight-line curve, the equation is represented as follows:

$y=mx+c$

Here, m is slope of the graph and c is intercept of the graph.

Answer to Problem 30.12QAP

The measured area $x=\frac{y-0.027}{0.8351}$

Explanation of Solution

The equation obtained from graph is $y=0.8351x+0.027$

And the equation of the line is $y=mx+c$

Where,

y is represented as y − intercept and the x is represented as x- intercept

m = slope of line

c = constant value

The rearrangement is done to convert in terms of measured area which is represented by the x −intercept or by x, and it arrives as,

$x=\frac{y-0.027}{0.8351}$

Interpretation Introduction

(c)

Interpretation:

The LOD in terms of moles needs to be determined, if the LOD for DOX is $3\text{x}{10}^{-11}$ M keeping the injection volume as 100 pL.

Concept introduction:

Laser induced fluorescence is spectroscopic method in which a molecule is excited to higher energy level by absorbing light. It helps to study the structure of molecule.

Answer to Problem 30.12QAP

LOD in moles is $\text{3x}{10}^{-21}$.

Explanation of Solution

LOD = $3\text{x}{10}^{-11}$

V (pL) = 100

V (L) = $\text{1x}{10}^{-10}$

The LOD in terms of moles is calculated by dividing the LOD value with injection volume in liter. This is given as-

$\text{moles=}\frac{\text{LOD}}{\text{volume(L)}}$

$\text{moles}=\frac{3\text{x}{10}^{-11}}{\text{1x}{10}^{-10}}=3\text{x}{10}^{-21}$

Hence, LOD (in moles) = $\text{3x}{10}^{-21}$

Interpretation Introduction

(d)

Interpretation:

The concentrations and standard deviations of the unknown DOX injected needs to be determined, if the peaks of two DOX unknown samples injected is 11.3 and 6.97.

Concept introduction:

The laser induced fluorescence is method wherein atoms or molecules is excited by passing laser light that excites them to higher energy levels releasing light spontaneously. It is useful in studying the structure of molecules.

Answer to Problem 30.12QAP

Standard deviation of peak 1 is 0.91 and standard deviation of peak 2 is 0.92.

Explanation of Solution

The graphical representation is −

Peak 1 = 11.3

Peak 2 = 6.97

Concentration of peak 1 is 13.49956.

Standard deviation is 0.90802029.

Concentration of peak 2 is 8.314343.

Standard deviation is 0.92408899.

Interpretation Introduction

(e)

Interpretation:

The time required for applied voltage to be doubled needs to be determined, if the DOX concentration is 300s. Also, the time needs to be determined, if the capillary length is doubled.

Concept introduction:

The laser induced fluorescence is method wherein atoms or molecules is excited though passing a laser light that excites them to higher energy levels releasing light spontaneously. It is useful in studying the structure of molecules.

Answer to Problem 30.12QAP

The time is halved if voltage is doubled. Its value is 150s.

The time is doubled if capillary length is doubled. Its value is 600s.

Explanation of Solution

Consider the equation,

$t=\frac{L{L}_t}{{\mu }_{tot}V}$

Where L=length of detector, L_t is total length, ${\mu }_{tot}$ is total mobility, V is voltage.

From the equation, it is seen that time and voltage are inversely proportional. Hence, if the voltage is increased two times, then the time is halved.

Given, time is 300s, and then Voltage is 150s.

From the equation it is seen that time and lengths are directly proportional. Hence, if length is doubled, then time is also doubled.

Given 300 s and if it is doubled it becomes, 600s, therefore the time is 600s.

Interpretation Introduction

(f)

Interpretation:

The value of N if the capillary length is doubled at the same voltage and the value of N if the applied voltage is doubled with same capillary length.

Concept introduction:

The laser induced fluorescence involves excitation of molecules or atoms to higher energy levels releasing light spontaneously through passing of laser light. It is useful in studying the structure of molecules.

Answer to Problem 30.12QAP

The value of N is 50,000 if the capillary length is doubled at same voltage.

The value of N is 200,000 if the voltage is doubled at same capillary length.

Explanation of Solution

Consider equation,

$N=\frac{{\mu }_{tot}V{L}_{}}{2D{L}_t}$

Where ${\mu }_{tot}$ is total mobility, V is voltage, L is length of detector, D is diffusion coefficient, and L_t is total length.

Given, plate count as 100,000.

From the given equation, the value of N is inversely proportional to total length. Hence, if the capillary length is doubled, the plate count is halved.

Therefore, N = 50,000.

Given plate count as 100,000.

From the given equation, the value of N is directly proportional to voltage; hence if voltage is doubled the value of N would also double. Hence, N =200,000.

Interpretation Introduction

(g)

Interpretation:

The plate height for capillary with N=100,000 for 40.6cm long capillary having diameter of 50µm needs to be determined.

Concept introduction:

The laser induced fluorescence involves excitation of molecules or atoms to higher energy levels releasing light spontaneously through passing of laser light. It is useful in studying the structure of molecules.

Answer to Problem 30.12QAP

The plate height is $4.06\text{x}{10}^{-4}\text{cm}$.

Explanation of Solution

Consider equation,

$H=\frac{L_t}{N}$

Where L_t is total length and N is the number of theoretical plates.

Given value of L_t as 40.6 cm and N as 100,000. Substitute in equation above,

$\begin{array}{c}H={\frac{L_t}{N}}_{}\\ =\frac{40.6cm}{100000}\\ =4.06\text{x}{10}^{-4}\end{array}$

Hence, the value of H is $4.06\text{x}{10}^{-4}\text{cm}$.

Interpretation Introduction

(h)

Interpretation:

The value of variance ${\sigma }^2$ needs to be determined, if the plate height for capillary with N=100,000 is 40.6cm having diameter of 50µm.

Concept introduction:

The laser induced fluorescence involves excitation of molecules or atoms to higher energy levels releasing light spontaneously through passing a laser light. It is useful in studying the structure of molecules.

Answer to Problem 30.12QAP

The value of ${\sigma }^2$ is $1.65\text{ x }{10}^{-2}\text{cm}$.

Explanation of Solution

Consider equation,

${\sigma }^2=H{L}_t$

Where L_t is total length, H is the plate height and ${\sigma }^2$ is variance.

Given value of L_t as 40.6 cm and H is $4.06\text{x}{10}^{-4}\text{cm}$. Substitute in equation above,

$\begin{array}{c}{\sigma }^2=H{L}_t\\ =4.06\times {10}^{-4}\text{cm x 40}\text{.6 cm}\\ \text{=165}\times {10}^{-4}\mathrm{cm}\\ =1.65\times {10}^{-2}\mathrm{cm}\end{array}$

Hence, the value of ${\sigma }^2$ is $1.65\times {10}^{-2}\text{cm}$.