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\(\bullet\) (a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength 0.0600 nm? (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength 0.0600 nm? (Hint: In each case the initial kinetic energy is negligible.)

Short Answer

Expert verified
The accelerating voltage needed for electrons is approximately 150.6 kV and for protons approximately 144.9 MV.

Step by step solution

01

Understand the De Broglie Wavelength

The de Broglie wavelength formula connects the wavelength \( \lambda \) of a particle to its momentum \( p \) by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \) J·s.
02

Relate Wavelength to Particle's Velocity

For a particle with mass \( m \) and velocity \( v \), momentum \( p = mv \). Therefore, the de Broglie wavelength equation becomes \( \lambda = \frac{h}{mv} \). Rearrange to find \( v = \frac{h}{m\lambda} \).
03

Determine Kinetic Energy

The kinetic energy \( KE \) of a particle is given by \( KE = \frac{1}{2}mv^2 \). Substitute velocity \( v = \frac{h}{m\lambda} \) into \( KE \), resulting in \( KE = \frac{1}{2}m \left( \frac{h}{m\lambda} \right)^2 \). Simplify to \( KE = \frac{h^2}{2m\lambda^2} \).
04

Convert Kinetic Energy to Accelerating Voltage

The kinetic energy gained by a charged particle under an accelerating voltage \( V \) is \( KE = eV \), where \( e \) is the elementary charge \( 1.602 \times 10^{-19} \) C. Equate \( eV = \frac{h^2}{2m\lambda^2} \) to solve for \( V \): \( V = \frac{h^2}{2me\lambda^2} \).
05

Calculate for Electrons

Using \( m_e = 9.11 \times 10^{-31} \) kg for the electron mass and \( \lambda = 0.0600 \times 10^{-9} \) m, find \( V = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-19} \times (0.0600 \times 10^{-9})^2} \). Calculate this to find: \( V \approx 150.6 \) kV.
06

Calculate for Protons

Using \( m_p = 1.67 \times 10^{-27} \) kg for the proton mass and the same \( \lambda \), substitute in \( V = \frac{(6.626 \times 10^{-34})^2}{2 \times 1.67 \times 10^{-27} \times 1.602 \times 10^{-19} \times (0.0600 \times 10^{-9})^2} \). Calculate to find: \( V \approx 144.9 \times 10^3 \) kV or \( V \approx 144.9 \) MV.

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Key Concepts

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

Accelerating Voltage
In an electron microscope, accelerating voltage plays a crucial role. It provides the energy needed to speed up electrons to high velocities. These high speeds enable electrons to exhibit smaller wavelengths, enhancing the resolution of the microscope. To achieve this, electrons are exposed to an electric field created by the accelerating voltage. This electric field increases the kinetic energy of the electrons from zero, as they are initially considered to have negligible kinetic energy.

The equation connecting the kinetic energy and accelerating voltage is\[ KE = eV \]where:- \( KE \) is the kinetic energy- \( e \) is the elementary charge \( (1.602 \times 10^{-19} \text{ C}) \)- \( V \) is the accelerating voltage

This relationship shows that as voltage increases, so does the kinetic energy and thus, the velocity of the electrons. This principle is fundamental in calculating the voltage required to attain specific wavelengths, like in our exercise, where electrons and protons need to reach a wavelength of 0.0600 nm.
de Broglie Wavelength
The de Broglie wavelength is a concept that reveals the wave-like nature of particles, including electrons. It's central to understanding how electron microscopes function. According to de Broglie, every moving particle or object has a wave associated with it. The wavelength of this wave, known as the de Broglie wavelength, is given by the equation:\[ \lambda = \frac{h}{p} \]where:- \( \lambda \) is the wavelength- \( h \) is Planck's constant \((6.626 \times 10^{-34} \text{ J·s})\)- \( p \) is the momentum of the particle

If we take momentum \( p \) as mass \( m \) times velocity \( v \), this can be rewritten as:\[ \lambda = \frac{h}{mv} \]

This expression relates a particle's wavelength directly to its speed and mass. As speed increases, the de Broglie wavelength decreases, allowing for greater resolving power in microscopes. When working with very small objects like electrons or protons, using high velocities reduces their wavelengths to values that enable them to resolve even more delicate structures.
Kinetic Energy
Kinetic energy is the energy a particle possesses due to its motion. In the context of electron or proton movement within an electron microscope, kinetic energy becomes significant. The particle's kinetic energy, once accelerated by the voltage, is expressed using:\[ KE = \frac{1}{2}mv^2 \]where:- \( m \) is the mass of the particle- \( v \) is its velocity

Rearranging this equation with de Broglie's expression for velocity, we obtain:\[ KE = \frac{h^2}{2m\lambda^2} \]

This equation highlights the inverse relationship between kinetic energy and the square of the de Broglie wavelength. When the wavelength is reduced, kinetic energy must increase, enhancing particle velocity. In calculating the kinetic energy necessary for a given wavelength, this equation shows that the effect of particle mass is also significant, differing for electrons and protons.
Elementary Charge
The concept of elementary charge is essential in calculating the energy dynamics within an electron microscope. The elementary charge, noted as \( e \), is a fundamental physical constant representing the electric charge carried by a single proton (or the negative of that carried by a single electron).\[ e = 1.602 \times 10^{-19} \text{ C} \]

The elementary charge is pivotal in linking potential difference or accelerating voltage with kinetic energy gain. In the equation:\[ KE = eV \]\( e \) acts as a conversion factor, turning voltage into kinetic energy. This conversion is principle in scenarios where particles like electrons or protons need to be accelerated to achieve certain wavelengths. Understanding the elementary charge helps in studying how charged particles behave under electrostatic influence and in various applications involving electric fields.

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Most popular questions from this chapter

\(\bullet\) A certain atom has an energy level 3.50 eV above the ground state. When excited to this state, it remains \(4.0 \mu s,\) on the average, before emitting a photon and returning to the ground state. (a) What is the energy of the photon? What is its wavelength? (b) What is the smallest possible uncertainty in energy of the photon?

\(\bullet\) (a) An electron moves with a speed of \(4.70 \times 10^{6} \mathrm{m} / \mathrm{s}\) . What is its de Broglie wavelength? (b) A proton moves with the same speed. Determine its de Broglie wavelength.

\(\bullet\) (a) The uncertainty in the \(x\) component of the position of a proton is \(2.0 \times 10^{-12} \mathrm{m}\) . What is the minimum uncertainty in the \(x\) component of the velocity of the proton? (b) The uncertainty in the \(x\) component of the velocity of an electron is 0.250 \(\mathrm{m} / \mathrm{s} .\) What is the minimum uncertainty in the \(x\) coordinate of the electron?

\(\bullet\) A specimen of the microorganism Gastropus hyptopus measures 0.0020 \(\mathrm{cm}\) in length and can swim at a speed of 2.9 times its body length per second. The tiny animal has a mass of roughly \(8.0 \times 10^{-12} \mathrm{kg}\) . (a) Calculate the de Broglie wave- length of this organism when it is swimming at top speed. (b) Calculate the kinetic energy of the organism (in eV) when it is swimming at top speed.

. An electron in an excited state of hydrogen makes a transition from the \(n=5\) level to the \(n=2\) level. (a) Does the atom emit or absorb a photon during this process? How do you know? (b) Calculate the wavelength of the photon involved in the transition.

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