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Why can photomultiplier tubes not be used with IR radiation?

Short Answer

Expert verified

A photon with energy in the ultraviolet-visible region is required to liberate electrons from the photo emissive surface of a photomultiplier tube.

Step by step solution

01

Step 1. Given information

Photomultiplier tubes are very sensitive light detectors that operate in the ultraviolet, visible, and near-infrared spectra. The current created by incident light is multiplied by these detectors.

02

Step 2. Infrared light does not require photomultiplier tubes

The power of infrared radiation is measured using a heat detector. This amount of infrared energy is insufficient to liberate electrons from the photomultiplier tube's photoemissive surface. A photon with energy in the ultraviolet-visible region is required to liberate electrons from the photo emissive surface of a photomultiplier tube.

As a result, infrared light does not require the usage of a photomultiplier tube.

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Most popular questions from this chapter

At 580 nm, which is the wavelength of its maximum absorption, the complex Fe(SCN)2+ has a molar absorptivity of 7.00 * 103 L cm -1 mol -1. Calculate

(a) the absorbance of a 4.47 * 10 -5M solution of the complex at 580 nm in a 1.00-cm cell.

(b) the absorbance of a solution in a 2.50-cm cell in which the concentration of the complex is one half that

in (a).

(c) the percent transmittance of the solutions described in (a) and (b).

(d) the absorbance of a solution that has half the transmittance of that described in (a).

A portable photometer with a linear response to radiation registered 56.3ยตA with the solvent in the light path. The photometer was set to zero with no light striking the detector. Replacement of the solvent with an absorbing solution yielded a response of 36.7ยตA. Calculate

(a) the percent transmittance of the sample solution.

(b) the absorbance of the sample solution.

(c) the transmittance to be expected for a solution in which the concentration of the absorber is one third that of the original sample solution.

(d) the transmittance to be expected for a solution that has twice the concentration of the sample solution.

A solution containing 3.92mg/100mLof A(335g/mol)has a transmittance of64.1% in a 1.50-cm cell at 425nm. Calculate the molar absorptivity of A at this wavelength.

The following questions concern the relative concentration uncertainty in spectrophotometry.

(a) If the relative concentration uncertainty is given by Equation 13-13, use calculus to show that the minimum uncertainty occurs at 36.8%T. What is the absorbance that minimizes the concentration uncertainty? Assume that sTis independent of concentration.

(b) Under shot-noise-limited conditions, the relative concentration uncertainty is given by Equation 13-14. Another form of the equation for the shot-noise-limited case is 13

scc=-kT-1/2InT

where kis a constant. Use calculus and derive the transmittance and absorbance that minimize the concentration uncertainty.

(c) Describe how you could experimentally determine whether a spectrophotometer was operating under Case I, Case II, or Case III conditions.

Calculate the absorbance of solutions having twice the percent transmittance of those in Problem 13-2.

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