/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 59 The photomultiplier tube in a co... [FREE SOLUTION] | 91Ó°ÊÓ

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The photomultiplier tube in a commercial scintillation counter contains 15 of the special electrodes, or dynodes. Each dynode produces 3 electrons for every electron that strikes it. One photoelectron strikes the first dynode. What is the maximum number of electrons that strike the 15th dynode?

Short Answer

Expert verified
14348907 electrons strike the 15th dynode.

Step by step solution

01

Understanding the Multiplication at Each Dynode

Each dynode in the photomultiplier tube multiplies the number of incoming electrons by a factor of 3. Therefore, to determine the increase in the number of electrons through each dynode, consider that if 1 electron hits a dynode, 3-electron output is produced.
02

Constructing the Exponential Relationship

Given that each dynode outputs 3 electrons for every input electron, the number of electrons hitting the next dynode is 3 times the number of electrons hitting the previous one. Since this occurs over 15 stages, we need to calculate the total electrons at the 15th dynode using exponential growth.
03

Calculating the Electron Count at the 15th Dynode

The process starts with 1 electron and is multiplied by 3 at each of the 15 dynodes. Therefore, the number of electrons reaching the 15th dynode is calculated as:\[ 3^{15} \]
04

Evaluating the Final Expression

Calculate the value of \(3^{15}\):\[ 3^{15} = 3 imes 3 imes 3 imes ext{...(15 times)} = 14348907 \]
05

Conclusion on Electron Multiplication

Thus, the maximum number of electrons that strike the 15th dynode is 14348907.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dynode multiplication
Dynode multiplication is a key process in the operation of photomultiplier tubes. These devices are commonly used in scientific instruments to detect very low levels of light.
In a photomultiplier tube, incoming light photons are first converted into photoelectrons. These photoelectrons then interact with specially designed electrodes called dynodes. Each dynode increases, or multiplies, the number of electrons. In our example, each dynode produces three electrons for every electron that hits it.
This multiplication process continues as the electrons move through subsequent dynodes. With each stage, the total number of electrons grows significantly, culminating in a cascade of electrons at the final dynode that is much larger than the initial number.
This cascading multiplication effect is pivotal for detecting faint signals, allowing small amounts of light to be amplified and measured accurately.
Exponential growth in physics
Exponential growth in physics describes a process where quantities increase rapidly, doubling or tripling at consistent intervals.
In the context of photomultiplier tubes, exponential growth occurs in the electron multiplication process across dynodes.
With each dynode tripling the number of electrons, the overall process is modeled by exponential growth. The formula used to calculate the number of electrons at the final dynode is based on the formula for exponential growth: \[ 3^n \]where \( n \) is the number of dynodes. In the provided solution, \( n = 15 \), leading to a significant increase in electrons from initially one photoelectron to a multitude, precisely calculated as 14,348,907 electrons.
Such powerful exponential growth enables the effective amplification necessary for the precision of photodetecting devices.
Electron multiplication
Electron multiplication is a fascinating phenomenon that exploits the principles of electron dynamics and amplification.
The goal of electron multiplication in devices like photomultiplier tubes is to detect and measure tiny electron counts, stemming from minimal initial inputs.
This is achieved by using a sequence of dynodes, each acting as an amplifier. When a single electron strikes a dynode, it triggers the release of multiple electrons. This secondary electron emission process is finely controlled and specifically tuned to produce a desired multiplication factor—in our studied case, a factor of three.
The cumulative effect of this multiplication across several dynodes can elevate a single photoelectron into millions of electrons, making unnoticeable signals measurable.
Through this meticulous electron multiplication process, photomultiplier tubes serve as highly sensitive detectors, crucial in areas like medical imaging, scientific research, and radiation measurement.

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Most popular questions from this chapter

A sample has a \(_{6}^{14} \mathrm{C}\) activity of \(0.0061 \mathrm{Bq}\) per gram of carbon. (a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq. (b) Evidence suggests that the value of \(0.23 \mathrm{Bq}\) might have been as much as \(40 \%\) larger. Repeat part (a), taking into account this \(40 \%\) increase.

Two radioactive waste products from nuclear reactors are stron\(\operatorname{tium}_{38}^{90} \mathrm{Sr}\left(T_{1 / 2}=29.1 \mathrm{yr}\right)\) and cesium \({ }_{35}^{134} \mathrm{Cs}\left(T_{1 / 2}=2.06 \mathrm{yr}\right) .\) These two species are present initially in a ratio of \(N_{0.5 \mathrm{r}} / N_{0 . \mathrm{Cs}}=7.80 \times 10^{-3} .\) What is the ratio \(N_{\mathrm{Sr}} / N_{\mathrm{Cs}}\) fifteen years later?

Two radioactive nuclei A and B are present in equal numbers to begin with. Three days later, there are three times as many A nuclei as there are \(\mathrm{B}\) nuclei. The half-life of species \(\mathrm{B}\) is 1.50 days. Find the halflife of species A.

To see why one curie of activity was chosen to be \(3.7 \times 10^{10} \mathrm{Bq}\) , determine the activity (in disintegrations per second) of one gram of radium \({ }_{88}^{226} \mathrm{Ra}\left(T_{1 / 2}=1.6 \times 10^{3} \mathrm{yr}\right)\).

Determine the symbol \({ }_{Z}^{A} \mathrm{X}\) for the parent nucleus whose \(\alpha\) decay produces the same daughter as the \(\beta^{-}\) decay of thallium \({ }_{81}^{208} \mathrm{Tl}\).

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