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

In the second type of helium-ion microscope, a 1.2-MeV ion passing through a cell loses 0.2 MeV per \(\mu\)m of cell thickness. If the energy of the ion can be measured to 6 keV, what is the smallest difference in thickness that can be discerned? (a) 0.03 \(\mu\)m; (b) 0.06 \(\mu\)m; (c) 3 \(\mu\)m; (d) 6 \(\mu\)m.

Short Answer

Expert verified
The smallest thickness difference discernible is 0.03 \(\mu\text{m}\) (option a).

Step by step solution

01

Identify the Problem

We need to find the smallest thickness difference that can be detected, given that the energy loss per micrometer (\(\mu\text{m}\)) is 0.2 MeV and the smallest measurable energy difference is 6 keV.
02

Convert Units

First, convert the smallest measurable energy difference from keV to MeV. Since 1 MeV = 1000 keV, 6 keV = 0.006 MeV.
03

Apply the Relation Between Energy Loss and Thickness

The energy loss per unit thickness is 0.2 MeV per \(\mu\text{m}\). Therefore, the equation relating energy loss \(\Delta E\) to thickness \(\Delta x\) is \(\Delta E = 0.2 \times \Delta x\).
04

Solve for Thickness Difference

Rearrange the equation to solve for thickness change: \(\Delta x = \frac{\Delta E}{0.2}\). Substitute \(\Delta E = 0.006\) MeV into the equation to get \(\Delta x = \frac{0.006}{0.2} = 0.03 \mu\text{m}\).

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!

Key Concepts

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

Energy Loss
In the context of helium-ion microscopy, energy loss refers to the reduction in energy of helium ions as they pass through a material. This phenomenon occurs due to interactions between the ions and atoms in the material.
These interactions can cause the ions to lose energy through processes such as electron excitation and atom displacements. In our example, a helium ion passing through a cell experiences an energy loss of 0.2 MeV per micrometer of cell thickness.
Understanding energy loss is crucial, as it impacts the resolution and effectiveness of the helium-ion microscope. This loss allows for precise measurements of material thickness and helps visualize structures at a microscopic level.
Micrometer Measurement
In scientific measurements, particularly in helium-ion microscopy, precision is key. The micrometer (μm) is a unit of length in the metric system, equal to one millionth of a meter. This tiny unit is perfect for measuring the thickness of cells or layers in various samples.
In our problem, it's crucial to determine how thick a sample must be for the energy loss to be detectable. Since the energy loss is tied closely to the thickness of the cell, micrometers serve as an ideal unit for these small-scale measurements.
This capability allows scientists to discern incredibly small differences in thickness, enhancing the accuracy and applicability of the helium-ion microscope in scientific research.
Energy Conversion
In the energy measurements of helium ions, conversion between different units is essential. An understanding of energy conversion helps bridge quantities, making the analysis accurate and meaningful.
Here, the energy is initially measured in keV (kiloelectronvolts) and is converted to MeV (megaelectronvolts) to match the units used for energy loss calculations. Recall that 1 MeV is 1000 times greater than 1 keV. Thus, when needing a precise measurement like 6 keV, conversion to 0.006 MeV is necessary.
This conversion ensures consistency across calculations, providing reliable results, and helps assure that energies are precisely compared and evaluated.
Ion Energy Measurement
Ion energy measurement is vital to exploit the full analytical power of the helium-ion microscope. Accurate measurement of energy helps detect small differences in material structures and thicknesses.
The helium-ion microscope allows measurement down to the keV scale, detecting very small changes in ion energies. This sensitivity ensures high-resolution imaging, contributing to the detailed insights available from this technology.
With such precise capability, it becomes possible to measure energy discrepancies that correlate with minute differences in thickness. This precision is imperative for applications ranging from materials science to biology, where details at the cellular and molecular level can be both subtle and significant.

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

To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. (a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? (b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

Determine \(\lambda_m\) , the wavelength at the peak of the Planck distribution, and the corresponding frequency \(f\), at these temperatures: (a) 3.00 K; (b) 300 K; (c) 3000 K.

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature \(T\) and a diameter 3.0 times that of the hotter star. (a) What is the temperature of the hotter star in terms of \(T\) ? (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

Radiation has been detected from space that is characteristic of an ideal radiator at \(T\) = 2.728 K. (This radiation is a relic of the Big Bang at the beginning of the universe.) For this temperature, at what wavelength does the Planck distribution peak? In what part of the electromagnetic spectrum is this wavelength?

An electron beam and a photon beam pass through identical slits. On a distant screen, the first dark fringe occurs at the same angle for both of the beams. The electron speeds are much slower than that of light. (a) Express the energy of a photon in terms of the kinetic energy \(K\) of one of the electrons. (b) Which is greater, the energy of a photon or the kinetic energy of an electron?

See all solutions

Recommended explanations on Physics 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