Harr Clinical Chemistry Exam Questions Save
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Terms in this set (620)
D. All of these options
D Absorbance is proportional to the inverse log of transmittance.
A = -log T = log 1/T
Which formula correctly Multiplying the numerator and denominator
describes the relationship by 100 gives:
between absorbance and %T ? A = log (100/100 X T) More
100 X T = %T,
A. A = 2 - log %T substituting %T for 100 X T gives:
B. A = log 1/T A = log 100/%T
C. A = -log T A = log 100 - log %T
D. All of these options A = 2.0 - log %T
For example, if %T = 10.0, then:
A = 2.0 - log 10.0
log 10.0 = 1.0
A = 2.0-1.0 = 1.0
A solution that has a B. 2.0
transmittance of 1.0 %T would
have an absorbance of: B
A = 2.0 - log %T
A. 1.0 A = 2.0 - log 1.0
B. 2.0 The log of 1.0 = 0
C. 1% A = 2.0
D. 99%
In absorption D. Absorbance is directly proportional to concentration
spectrophotometry:
D Beer's law states that A = a × b × c, where a is the absorptivity
A. Absorbance is directly coefficient (a constant), b is the path length, and c is
proportional to concentration. Absorbance is directly proportional to both b and
transmittance c. Doubling the path length results in incident light contacting
B. Percent transmittance is twice the number of molecules in solution. This causes
directly proportional to absorbance to double, the same effect as doubling the
concentration concentration of molecules.
C. Percent transmittance is
directly proportional to the light
path length
D. Absorbance is directly
proportional to concentration
,Which wavelength would be A. 450 nm
absorbed strongly by a red-
colored solution? A A solution transmits light corresponding in wavelength to its
color, and usually absorbs light of wavelengths complementary to
A. 450 nm its color. A red solution transmits light of 600-650 nm and strongly
B. 585 nm absorbs 400-500 nm light.
C. 600 nm
D. 650 nm
B. 525 nm
A green-colored solution would
show highest transmittance at:
B Green light consists of wavelengths from 500-550 nm. A green-
colored solution with a transmittance maximum of 525 nm and a
A. 475 nm
50-nm bandpass transmits light of 525 nm and absorbs light
B. 525 nm
below 475 nm and above 575 nm. A solution that is green would
C. 585 nm
be quantitated using a wavelength that it absorbs strongly, such as
D. 620 nm
450 nm.
SITUATION: A technologist is A. Replace the source lamp
performing an enzyme assay at
340 nm using a visible-range A Visible spectrophotometers are usually supplied with a tungsten
spectrophotometer. After setting or quartz halogen source lamp. Tungsten lamps produce a
the wavelength and adjusting the continuous range of wavelengths from about 320-2,000 nm.
readout to zero %T with the light Output increases as wavelength becomes longer peaking at
path blocked, a cuvette with around 1,000 nm, and is poor below 400 nm. As the lamp
deionized water is inserted. With envelope darkens with age, the amount of light reaching the
the light path fully open and the photodetector at 340 nm becomes insufficient to set the blank
100%T control at maximum, the reading to 100%T. Quartz halogen lamps produce light from 300
instrument readout will not rise nm through the infrared region. Deuterium or hydrogen lamps
above 90%T. What is the most produce ultraviolet-rich spectra optimal for ultraviolet (UV) work.
appropriate first course of Mercury vapor lamps produce a discontinuous spectrum that
action? includes a high output at around 365 nm that is useful for
fluorescent applications. Xenon lamps generate a continuous
A. Replace the source lamp spectrum of fairly uniform intensity from 300-2,000 nm, making
B. Insert a wider cuvette into the them useful for both visible and UV applications.
light path
C. Measure the voltage across
the lamp terminals
D. Replace the instrument fuse
, Which type of monochromator D. A prism and a variable exit slit
produces the purest
monochromatic light in the UV D Diffraction gratings and prisms both produce a continuous
range? range of wavelengths. A diffraction grating produces a uniform
separation of wavelengths. A prism produces much better
A. A diffraction grating and a separation of high-frequency light because refraction is greater
fixed exit slit for higher-energy wavelengths. Instruments using a prism and a
B. A sharp cutoff filter and a variable exit slit can produce UV light of a very narrow bandpass.
variable exit slit The adjustable slit is required in order to allow sufficient light to
C. Interference filters and a reach the detector to set 100%T.
variable exit slit
D. A prism and a variable exit slit
Which monochromator D. 5-nm bandpass
specification is required in order
to measure the true absorbance D Bandpass refers to the range of wavelengths passing through
of a compound having a natural the sample. The narrower the bandpass, the greater the
absorption bandwidth of 30 nm? photometric resolution. Bandpass can be made smaller by
reducing the width of the exit slit. Accurate absorbance
A. 50-nm bandpass measurements require a bandpass less than one-fifth the natural
B. 25-nm bandpass bandpass of the chromophore
C. 15-nm bandpass
D. 5-nm bandpass
D. Photomultiplier tube
Which photodetector is most
sensitive to low levels of light?
D The photomultiplier tube uses dynodes of increasing
voltage to amplify the current produced by the photosensitive
A. Barrier layer cell
cathode. It is 10,000 times as sensitive as a barrier layer cell, which
B. Photodiode
has no amplification. A photomultiplier tube requires a DC-
C. Diode array
regulated lamp because it responds to light fluctuations caused
D. Photomultiplier tube
by the AC cycle.
Which condition is a common C. Dispersion from second-order spectra
cause of stray light?
C Stray light is caused by the presence of any light other than the
A. Unstable source lamp voltage wavelength of measurement reaching the detector. It is most
B. Improper wavelength often caused by second-order spectra, deteriorated optics, light
calibration dispersed by a darkened lamp envelope, and extraneous room
C. Dispersion from second-order light.
spectra
D. Misaligned source lamp
with Verified Solutions
Terms in this set (620)
D. All of these options
D Absorbance is proportional to the inverse log of transmittance.
A = -log T = log 1/T
Which formula correctly Multiplying the numerator and denominator
describes the relationship by 100 gives:
between absorbance and %T ? A = log (100/100 X T) More
100 X T = %T,
A. A = 2 - log %T substituting %T for 100 X T gives:
B. A = log 1/T A = log 100/%T
C. A = -log T A = log 100 - log %T
D. All of these options A = 2.0 - log %T
For example, if %T = 10.0, then:
A = 2.0 - log 10.0
log 10.0 = 1.0
A = 2.0-1.0 = 1.0
A solution that has a B. 2.0
transmittance of 1.0 %T would
have an absorbance of: B
A = 2.0 - log %T
A. 1.0 A = 2.0 - log 1.0
B. 2.0 The log of 1.0 = 0
C. 1% A = 2.0
D. 99%
In absorption D. Absorbance is directly proportional to concentration
spectrophotometry:
D Beer's law states that A = a × b × c, where a is the absorptivity
A. Absorbance is directly coefficient (a constant), b is the path length, and c is
proportional to concentration. Absorbance is directly proportional to both b and
transmittance c. Doubling the path length results in incident light contacting
B. Percent transmittance is twice the number of molecules in solution. This causes
directly proportional to absorbance to double, the same effect as doubling the
concentration concentration of molecules.
C. Percent transmittance is
directly proportional to the light
path length
D. Absorbance is directly
proportional to concentration
,Which wavelength would be A. 450 nm
absorbed strongly by a red-
colored solution? A A solution transmits light corresponding in wavelength to its
color, and usually absorbs light of wavelengths complementary to
A. 450 nm its color. A red solution transmits light of 600-650 nm and strongly
B. 585 nm absorbs 400-500 nm light.
C. 600 nm
D. 650 nm
B. 525 nm
A green-colored solution would
show highest transmittance at:
B Green light consists of wavelengths from 500-550 nm. A green-
colored solution with a transmittance maximum of 525 nm and a
A. 475 nm
50-nm bandpass transmits light of 525 nm and absorbs light
B. 525 nm
below 475 nm and above 575 nm. A solution that is green would
C. 585 nm
be quantitated using a wavelength that it absorbs strongly, such as
D. 620 nm
450 nm.
SITUATION: A technologist is A. Replace the source lamp
performing an enzyme assay at
340 nm using a visible-range A Visible spectrophotometers are usually supplied with a tungsten
spectrophotometer. After setting or quartz halogen source lamp. Tungsten lamps produce a
the wavelength and adjusting the continuous range of wavelengths from about 320-2,000 nm.
readout to zero %T with the light Output increases as wavelength becomes longer peaking at
path blocked, a cuvette with around 1,000 nm, and is poor below 400 nm. As the lamp
deionized water is inserted. With envelope darkens with age, the amount of light reaching the
the light path fully open and the photodetector at 340 nm becomes insufficient to set the blank
100%T control at maximum, the reading to 100%T. Quartz halogen lamps produce light from 300
instrument readout will not rise nm through the infrared region. Deuterium or hydrogen lamps
above 90%T. What is the most produce ultraviolet-rich spectra optimal for ultraviolet (UV) work.
appropriate first course of Mercury vapor lamps produce a discontinuous spectrum that
action? includes a high output at around 365 nm that is useful for
fluorescent applications. Xenon lamps generate a continuous
A. Replace the source lamp spectrum of fairly uniform intensity from 300-2,000 nm, making
B. Insert a wider cuvette into the them useful for both visible and UV applications.
light path
C. Measure the voltage across
the lamp terminals
D. Replace the instrument fuse
, Which type of monochromator D. A prism and a variable exit slit
produces the purest
monochromatic light in the UV D Diffraction gratings and prisms both produce a continuous
range? range of wavelengths. A diffraction grating produces a uniform
separation of wavelengths. A prism produces much better
A. A diffraction grating and a separation of high-frequency light because refraction is greater
fixed exit slit for higher-energy wavelengths. Instruments using a prism and a
B. A sharp cutoff filter and a variable exit slit can produce UV light of a very narrow bandpass.
variable exit slit The adjustable slit is required in order to allow sufficient light to
C. Interference filters and a reach the detector to set 100%T.
variable exit slit
D. A prism and a variable exit slit
Which monochromator D. 5-nm bandpass
specification is required in order
to measure the true absorbance D Bandpass refers to the range of wavelengths passing through
of a compound having a natural the sample. The narrower the bandpass, the greater the
absorption bandwidth of 30 nm? photometric resolution. Bandpass can be made smaller by
reducing the width of the exit slit. Accurate absorbance
A. 50-nm bandpass measurements require a bandpass less than one-fifth the natural
B. 25-nm bandpass bandpass of the chromophore
C. 15-nm bandpass
D. 5-nm bandpass
D. Photomultiplier tube
Which photodetector is most
sensitive to low levels of light?
D The photomultiplier tube uses dynodes of increasing
voltage to amplify the current produced by the photosensitive
A. Barrier layer cell
cathode. It is 10,000 times as sensitive as a barrier layer cell, which
B. Photodiode
has no amplification. A photomultiplier tube requires a DC-
C. Diode array
regulated lamp because it responds to light fluctuations caused
D. Photomultiplier tube
by the AC cycle.
Which condition is a common C. Dispersion from second-order spectra
cause of stray light?
C Stray light is caused by the presence of any light other than the
A. Unstable source lamp voltage wavelength of measurement reaching the detector. It is most
B. Improper wavelength often caused by second-order spectra, deteriorated optics, light
calibration dispersed by a darkened lamp envelope, and extraneous room
C. Dispersion from second-order light.
spectra
D. Misaligned source lamp
