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Chapter 20: Q33P (page 527)

A spectrum has a signal-to-noise ratio of 8/1. How many spectra must be averaged to increase the signal-to-noise ratio to 20 / 1?

Short Answer

Expert verified

The spectra must be averaged to increase the signal-to-noise ratio to 20 / 1 is 7 scans required.

Step by step solution

01

Definition of Signal to noise

In science and engineering, the signal-to-noise ratio compares the level of a desired signal to the amount of background noise.
The signal-to-noise ratio (SNR) is frequently given in decibels and is defined as the ratio of signal power to noise power.

02

Determine the many spectra must be averaged to increase the signal-to-noise ratio to 20 / 1

The spectra must be averaged in order to enhance the signal-to-noise ratio to 20 / 1.
Initial ratio is as follows: 8/1
The signal-to-noise ratio is increased by averaging the factor of a particular ratio 2.5
It must now raise the ratio from 8 to 20 (factor of $\frac{20}{8} = 2.5$ )
As a result, the necessary factor is $2 . 5^{2} = 6.25$ there are around 7 scans.
To boost the signal-to-noise ratio to 20 / 1, the spectra must be averaged, which takes 7 scans.

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Most popular questions from this chapter

(a) In the cavity ring-down measurement at the opening of this chapter, absorbance is given by $A = \frac{L}{cln 10} (\frac{1}{τ} - \frac{1}{τ_{0}})$ whereis the length of the triangular path in the cavity, Cis the speed of light, Tis the ring-down lifetime with sample in the cavity, and $T_{0}$ is the ring-down lifetime with no sample in the cavity. Ring-down lifetime is obtained by fitting the observed ring-down signal intensityto an exponential decay of the form $l = l_{0} e^{- yτ}$ , whereis the initial intensity and t is time. A measurement ofis made at a wavelength absorbed by the molecule. The ring-down lifetime for 21.0-cm-1 along empty cavity is $18.52 μs$ and $18.52 μs$ for a cavity containing.role="math" localid="1664865078479" ${CO}_{2}$ Find the absorbance of ${CO}_{2}$ at this wavelength.

(b) The ring-down spectrum below arises from ${}^{13}{CH}_{4}$ and ${}^{12}{CH}_{4}$ from $1.9 ppm (vol / yol)$ of methane in outdoor air at 0.13. The spectrum arises from individual rotational transitions of the ground vibrational state to a second excited C-H)vibrational state of the molecule. (i) Explain what quantity is plotted on the ordinate ( Y-axis). (ii) The peak for ${}^{12}{CH}_{4}$ is at $6046.9546 {cm}^{- 1}$ . What is the wavelength of this peak in? What is the name of the spectral region where this peak is found?

The true absorbance of a sample is 1.000, but the mono chromator passes 1.0% stray light. Add the light coming through the sample to the stray light to find the apparent transmittance of the sample. Convert this answer back into absorbance and find the relative error in the calculated concentration of the sample.

The path length of a cell for infrared spectroscopy can be measured by counting interference fringes (ripples in the transmission spectrum). The following spectrum shows 30interference maxima between 1906and $698 c m^{- 1}$ obtained by placing an empty KBrcell in a spectrophotometer.

The fringes arise because light reflected from the cell compartment interferes constructively or destructively with the unreflected beam.

If the reflected beam travels an extra distance, it will interfere constructively with the unreflected beam. If the reflection path length is $λ / 2$ , destructive interference occurs. Peaks therefore arise when $m λ / 2 = 2 b$ and troughs occur when, where $m λ / 2 = 2 b$ is an integer. If the medium between KBr theplates has refractive index n, the wavelength in the medium is l/n, so the equations become $m λ / n = 2 b a n d m λ / 2 n = 2 b$ . The cell path length can be shown to be given by

$b = \frac{N}{2 n} \times \frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}} = \frac{N}{2 n} \times \frac{1}{{\tilde{v}}_{1} - {\tilde{v}}_{2}}$

where Nmaxima occur between wavelengths $λ_{1}$ and $λ_{2}$ . Calculate the path length of the cell that gave the interference fringes shown earlier.

Describe the role of each component of the spectrophotometer.

Would you use a tungsten or a deuterium lamp as a source of 300-nm radiation? What kind of lamp provides radiation at 4-mm wavelength?

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