Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Explain how the following detectors of visible radiation work: (a) photomultiplier tube, (b) photodiode array, (c) charge coupled device, and (d) image intensified charge coupled device.

Short Answer

Expert verified

(a)Photomultiplier tube is a photosensitive electron tube

(b)A photodiode is a p-n junction.

(c)In charge coupled device the image is projected onto the sensor by a lens and falls on the photoactive part composed of a network of photodiodes behind which are the capacitors.

(d) Image intensified charge coupled device amplifies incoming photons by prior to detection

Step by step solution

01

Step: 1 Definition of visible radiation:

Visual radiation, also called visible light or “light,” is a form of electromagnetic radiation.

02

Step: 2 The photomultiplier tube:

(a) Photomultiplier tube is a photosensitive electron tube in which light by an external photoelectric effect, as in a photocell, expels electrons from the photosensitive cathode between the cathode and the anode there is a series of electrodes, the so-called dynodes, each at an ever-increasing positive potential. Accelerated primary electrons hit the dynodes and each erupts several secondary electrons, so the electron current is multiplied many times over.

03

Step: 3 The photodiode array:

(b)

A photodiode is a p-n junction. When a photon of sufficient energy strikes a diode, it creates an electron-hole pair. This mechanism is also known as the internal photoelectric effect. If absorption occurs at the junctions of the depletion region or one diffuse length thereof, these carriers are removed from the junction by the embedded electric field of the depletion region. Thus the holes move towards the anode and also the electrons towards the cathode and a photocurrent is made. The full current through the photodiode is that the total of the dark current and also the photocurrent, therefore the dark current should be reduced to extend the sensitivity of the device.

04

Step: 4 The charge coupled device:

In charge coupled device the image is projected onto the sensor by a lens and falls on the photoactive part composed of a network of photodiodes behind which are the capacitors. The more light falls on the photodiode, the more it is charged accumulates in the accompanying capacitor. After the entire net was exposed to light, the control circuit causes each capacitor to transfer its charge content to its neighbor. The last capacitor in the series switches the charges to an amplifier that converts them into voltage. By repeating this process, the control circuit converts the contents of the entire network in a series of voltages.

05

Step: 5 The image intensified charge coupled device:

Image intensified charge coupled device amplifies incoming photons by prior to detection. Photons enters the exhausted device through a window at the left and strikes a cathode that's optimized for ultraviolet, visible, or infrared wavelengths. Once the cathode is irradiated by light, it emits electrons. Electrons from the cathode strike the microchannel plate. The potential becomes a lot of positive from left to right in every channel within the enlargement. A single electron hanging a channel causes many secondary electrons to be emitted. Emitted electrons are accelerated by the a lot of positive potential deeper into the channel. Every secondary electron strikes the wall additional into the channel and emits many a lot of electrons. A microchannel behaves sort of a photomultiplier tube with continuous walls rather than separate plates. Electrons rising from the microchannel plates strike a fluorescent screen that emits photons that are carried through optical fibers onto the pixels of the charge coupled device.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Refer to the Fourier transform infrared spectrum in Figure 20-33.

(a) The interferogram was sampled at retardation intervals of1.2260×10-4cm. What is the theoretical wavenumber range (0 to ?) of the

spectrum?

(b) A total of 4 096 data points were collected from δ=-toδ=+. Compute the value of, the maximum retardation.

(c) Calculate the approximate resolution of the spectrum.

(d) The interferometer mirror velocity is given in the figure caption. How many microseconds elapse between each datum?

(e) How many seconds were required to record each interfero gram once?

(f) What kind of beam splitter is typically used for the region 400 to 4 000cm-1? Why is the region below 400cm-1not observed?

A measurement with a signal-to-noise ratio of 100 / 1 can be thought of as a signal, S, with 1 % uncertainty, e. That is, the measurement isS±e=100±1.

(a) Use the rules for propagation of uncertainty to show that, if you add two such signals, the result is total signal=200±2 , giving a signal-to-noise ratio of200/2=141/1 .

(b) Show that if you add four such measurements, the signal-tonoise ratio increases to 200 / 1.

(c) Show that averaging n measurements increases the signal-to-noise ratio by a factor ofn compared with the value for one measurement.

Explain how a laser generates light. List important properties of laser light.

Here is an extremely sensitive method for measuring nitrite () down to in natural waters. The water sample is treated with sulfanilamide and N-(1-naphthylethylenediamine) in acid solution to produce a colored product with a molar absorptivity of 4.5×104M-1cm-1at 540 nm. The colored solution is pumped into a 4.5-meter-long, coiled Teflon tube whose fluorocarbon wall has a refractive index of 1.29. The aqueous solution inside the tube has a refractive index near 1.33. The colored solution is pumped through the coiled tube. An optical fiber delivers white light into one end of the tube, and an optical fiber at the other end leads to a polychromator and detector.

(a) Explain the purpose of the coiled Teflon tube and explain how it functions.

(b) What is the critical angle of incidence for total internal reflection at the Teflon/water interface?

(c) What is the predicted absorbance of a1.0nM solution of colored reagent?

(a) What resolution is required for a diffraction grating to resolve wavelengths of 512.23 and 512.26 nm? (b) With a resolution of 104, how close in nm is the closest line to 512.23 nm that can barely be resolved? (c) Calculate the fourth-order resolution of a grating that is 8.00 cm long and is ruled at 185 lines/mm. (d) Find the angular dispersion (ϕ) between light rays with wavelengths of 512.23 and 512.26 nm for first-order diffraction (n 5 1) and thirtieth-order diffraction from a grating with 250 lines/mm and f 5 3.08.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free