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instrumental Problem_set_4 - -Yes since the absorbance is...

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Joanne Muriithi. Chem 4242 Problem set 4 1.The throughput/Jaquinot advantage is the amount of digital data per time unit that is delivered. FT instruments have few optical elements and no slits to attenuate radiation, therefore radiant power reaches that reaches is much greater. For the same resolution, the throughput can be higher in an interferometer and it has an ability to achieve the same signal-to-noise ratio as a dispersive instrument, but in a much shorter time. Multiplex Advantage - means that the Fourier Transform instrument simultaneously measures a signal over a range of frequencies and obtains a advantage in the time t required to obtain a given signal-to-noise ration is shorter compared to that which would be necessary using dispersive methods. 2a. - Yes, detection would still occur without the wavelength selector. - Emissions above 400 nm is a nuisance thus it is a limitation. - Refractive index is directly proportional to the concentration (c), therefore
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Unformatted text preview: -Yes, since the absorbance is constant (633 nm), Beer’s law is followed. 2b. 1. The purpose of Operational Amplifier A1 is to integrate the input signal in terms of time while the operational amplifier A2 amplifies the voltage. 2. 1mA of current is needed so as not to exceed the output of 10V 3. Low pass filter – is used to absorb the noise Voltage follower – prevents loading errors Current converter – maintains linearity in the photomultiplier Capacitor – it acts as a low pass filter and reduces the voltage Resistor – converts current in to voltage 4. The capacitor and resistors impede the amount of voltage that passes through the operational amplifier which affects the a.c response. 2c) 12 Bit: 10V/4096 = 0.0024 14 Bit: 10V/16384= 6.1 x 10-4 16 bit: 10V/65536 = 1.5 x 10-4 The 12 bit is better since it has a high resolution and it gives out a more precise number....
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