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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 \(\Delta x = \frac{0.006}{0.2} = 0.03 \mu\text{m}\).

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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.

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Most popular questions from this chapter

The radii of atomic nuclei are of the order of 5.0 \(\times\) 10\(^{-15}\) m. (a) Estimate the minimum uncertainty in the momentum of an electron if it is confined within a nucleus. (b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the kinetic energy of an electron confined within a nucleus. (c) Compare the energy calculated in part (b) to the magnitude of the Coulomb potential energy of a proton and an electron separated by 5.0 \(\times\) 10\(^{-15}\) m. On the basis of your result, could there be electrons within the nucleus? (\(Note\): It is interesting to compare this result to that of Problem 39.72.)

Consider a particle with mass m moving in a potential \(U = {1\over2} kx^2\), as in a mass-spring system. The total energy of the particle is \(E = (p^2/2m) + 12 kx^2\). Assume that \(p\) and \(x\) are approximately related by the Heisenberg uncertainty principle, so \(px \approx h\). (a) Calculate the minimum possible value of the energy \(E\), and the value of \(x\) that gives this minimum E. This lowest possible energy, which is not zero, is called the \(zero-point \space energy\). (b) For the \(x\) calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle?

A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 1.25 nm apart. You observe that the first double-slit interference dark fringe occurs at \(\pm\)18.0\(^\circ\) from the original direction of the beam when viewed on a distant screen. (a) Are these electrons relativistic? How do you know? (b) Through what potential difference were the electrons accelerated?

(a) A particle with mass \(m\) has kinetic energy equal to three times its rest energy. What is the de Broglie wavelength of this particle? (\(Hint\): You must use the relativistic expressions for momentum and kinetic energy: \(E^2 = (pc^2) + (mc^2)^2\) and \(K = E - mc^2\).) (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider a hydrogenlike atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in electron volts)? (c) What is the wavelength of the radiation emitted in the transition from the \(n\) = 2 level to the \(n\) = 1 level?

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