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In the thickness-sensitive ion microscope, a \(1.2 \mathrm{MeV}\) ion passing through a cell loses \(0.2 \mathrm{MeV}\) per \(\mu \mathrm{m}\) of thickness of the cell. If the energy of the ion can be measured to \(6 \mathrm{keV},\) what is the smallest difference in thickness that can be discerned? A. \(0.03 \mu \mathrm{m}\) B. \(0.06 \mu \mathrm{m}\) C. \(3 \mu \mathrm{m}\) D. \(6 \mu \mathrm{m}\)

Short Answer

Expert verified
The smallest difference in thickness that can be discerned is A. 0.03 \(\mu m\).

Step by step solution

01

Identify Given Data

We have the following data given in the problem: The initial energy of the ion is 1.2 MeV. The ion loses 0.2 MeV per micrometer of thickness as it passes through the cell. The measurement resolution for energy is 6 keV.
02

Convert Energy to Consistent Units

The energy loss is given in MeV (0.2 MeV per \(\mu m\)), while the measurement resolution is in keV (6 keV). We need to convert the 6 keV into MeV to ensure consistent units. Since 1 MeV = 1000 keV, then 6 keV = 0.006 MeV.
03

Set Up the Energy Loss Equation

The energy loss per unit thickness (\(T\)) of the cell is given by 0.2 MeV/\(\mu m\). We need to find out the smallest difference in thickness (\(\Delta T\)) that corresponds to the smallest measurable energy difference, which is 0.006 MeV.
04

Solve for Smallest Thickness Difference

Using the relationship \(\Delta E = \text{energy loss per } \mu m \times \Delta T\), solve for \(\Delta T\). We have: \[ \Delta T = \frac{\Delta E}{\text{energy loss per } \mu m} = \frac{0.006 \text{ MeV}}{0.2 \text{ MeV/}\mu m} \]\[\Delta T = 0.03 \mu m \]
05

Choose the Correct Answer

Based on our calculation, the smallest difference in thickness that can be discerned is \(0.03 \mu m\). Therefore, the correct answer is A. \(0.03 \mu m\).

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Key Concepts

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

Ion Microscope
An ion microscope is a unique instrument that scientists use to visualize ultra-small structures like cells. It operates on the principle of ion beams instead of light or electrons used in traditional microscopes. In an ion microscope, a focused beam of ions interacts with the sample, causing changes in energy that can be measured. This instrument enables researchers to achieve incredibly high resolution, unparalleled by other types of microscopy.

Ion microscopes are especially useful in research fields that require precise structural information, such as materials science, biology, and nanotechnology. Through detailed analysis of how ions lose energy during their journey through a sample, scientists can determine various properties, including material composition and thickness.

Here are a few advantages of using ion microscopes:
  • High Resolution: Provides detailed imaging possibilities down to the nanoscale.
  • Material Versatility: Works with a wide range of substances.
  • Analytical Capabilities: Allows material composition analysis alongside structural observation.
Energy Loss
Understanding energy loss is crucial in measuring sample properties with an ion microscope. As ions pass through a material, they lose energy primarily through interactions with atoms in the sample. This energy loss is informative; by measuring it, we can infer certain characteristics of the substance, such as its thickness.

In the context of the original exercise, we are told that ions lose energy at a rate of \(0.2 \, \text{MeV}\) per micrometer of thickness. This means for every micrometer an ion moves through, it sheds part of its energy. These interactions are subtle and depend on factors like ion speed and sample density.

Simplifying the concept:
  • The faster an ion moves, the more energy it can potentially lose per unit distance.
  • Ions transfer this energy to the sample, either via atoms' displacement or ionization.
  • Such energy loss needs to be meticulously monitored for precise property measurement.
Measurement Resolution
Measurement resolution refers to the smallest change in a quantity that an instrument can detect. In the context of an ion microscope, this refers to its ability to measure small changes in ion energy accurately. Good measurement resolution is crucial for detecting minor alterations in sample properties, which informs the consequent analysis.

For our ion microscope problem, the measurement resolution for energy is stated as 6 keV. This implies the microscope can distinguish energy changes as minute as 6 keV in the energy of the ions. Conversion to more common units of measurement (such as MeV) can help in aligning this resolution with designations of energy loss, providing a direct correlation between energy measurements and thickness measurements.

To consider measurement resolution:
  • Use consistent units for accuracy.
  • A higher resolution indicates finer distinctions between measurements.
  • Resolution impacts the precision of derived material properties, such as thickness.
Thickness Calculation
Determining the thickness of a sample using an ion microscope involves understanding the relationship between energy loss and measurement resolution. As explained, as ions travel through a thicker material, they lose more energy. This change is used to calculate the material's thickness.

Given the data in our exercise, we calculated the smallest discernible thickness. This calculation involves solving the equation of energy loss per thickness unit, set against the smallest measurable energy difference. Specifically, with an energy loss rate of \(0.2 \, \text{MeV}\) per micrometer, and a measurable energy difference of \(0.006 \, \text{MeV}\), we find the smallest thickness difference, \(abla T \), using:\[ abla T = \frac{0.006 \, \text{MeV}}{0.2 \, \text{MeV/μm}} = 0.03 \, \mu m\]By being able to precisely measure minor differences in thickness, researchers can accurately analyze the properties of their samples at a microscopic level. This capacity is invaluable for precise scientific applications and further research developments.

In essence, being proficient in such calculations allows scientists to:
  • Precisely determine sample characteristics.
  • Control and hone measurement techniques.
  • Advance research with specialized focus on nanoscale materials.

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Most popular questions from this chapter

(a) What is the minimum potential difference between the filament and the target of an X-ray tube if the tube is to accelerate electrons to produce X-rays with a wavelength of \(0.150 \mathrm{nm} ?\) (b) What is the shortest wavelength produced in an X-ray tube operated at \(30.0 \mathrm{kV} ?\) (c) Would the answers to parts (a) and (b) be different if the tube accelerated protons instead of electrons? Why or why not?

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