Question

(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

The accelerating voltage for electrons is approximately 104,000 V; for protons, it is 54.1 billion V.

Step-by-step solution

Understanding the De Broglie Wavelength

The de Broglie wavelength \( \lambda \) of a particle is given by the equation \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For particles accelerated by a voltage \( V \), the momentum \( p \) can be expressed in terms of their kinetic energy, which derives from the electrical potential energy.

Relating Wavelength to Voltage for Electrons

For electrons, the de Broglie wavelength \( \lambda = 0.0600 \times 10^{-9} \, \text{m} \). The momentum \( p \) is related to kinetic energy by \( p^2 = 2meKE \), where \( KE = eV \). Therefore, \( \lambda = \frac{h}{\sqrt{2meV}} \), where \( m \) is the electron mass and \( e \) is the elemental charge.

Calculate Accelerating Voltage for Electrons

Rearrange the equation \( \lambda = \frac{h}{\sqrt{2meV}} \) to find \( V \). Substituting \( \lambda = 0.0600 \times 10^{-9} \, \text{m} \), \( m = 9.109 \times 10^{-31} \, \text{kg} \), \( e = 1.602 \times 10^{-19} \, \text{C} \), and \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \), solve for \( V \). This gives approximately \( V = 104 \times 10^{3} \, \text{V} \).

Relating Wavelength to Voltage for Protons

For protons, use the same relation: \( \lambda = \frac{h}{\sqrt{2mpV}} \), where \( mp \) is the proton mass \( 1.673 \times 10^{-27} \, \text{kg} \). Substitute \( \lambda = 0.0600 \times 10^{-9} \, \text{m} \), along with known \( h \) and \( e \), into this expression.

Calculate Accelerating Voltage for Protons

Using the rearranged formula and substituting the appropriate values for protons, solve for \( V \). This results in approximately \( V = 54.1 \times 10^{9} \, \text{V} \).

Key concepts

Electron Microscope

An electron microscope is a powerful tool that allows us to see images of tiny objects, such as cells and bacteria, at incredibly high magnifications. Unlike light microscopes that use light to illuminate specimens, electron microscopes use a beam of electrons. This allows for much higher resolution since electrons have much shorter wavelengths than visible light.
Electron microscopes work by accelerating electrons towards the sample we are examining. Once the electrons hit the sample, they interact with it, providing a detailed image of its structure. The shorter the wavelength of the electrons, the more detailed the image. This is why the concept of de Broglie wavelength is critical—because it relates to how we control the wavelength of these electrons using voltage.

• Shorter wavelengths yield higher resolution images.
• Electrons are accelerated using high voltages.
• De Broglie wavelength helps us understand how to calculate the needed voltage for clarity.

Accelerating Voltage

Accelerating voltage is a crucial component in electron microscopy. It refers to the voltage used to speed up the electrons in the electron microscope. The greater the voltage, the faster the electrons are accelerated, reducing their wavelength in the process.
The relationship between voltage and electron speed is founded on the conversion of electric potential energy into kinetic energy. As electrons are released from a cathode, they are accelerated across a voltage difference. This process converts the potential energy (due to the voltage) directly into the electrons' kinetic energy.

• Higher accelerating voltage reduces the effective wavelength of electrons.
• This lower wavelength aids in achieving greater resolution in electron microscopes.

Understanding and calculating the right accelerating voltage is vital. It allows scientists to acquire clearer and more detailed images.

Kinetic Energy

Kinetic energy in the context of an electron microscope represents the energy that electrons gain as they are accelerated by the voltage applied. Initially, electrons have negligible kinetic energy, but upon acceleration, they gain significant kinetic energy due to their movement.
In our equations, the kinetic energy is related to the voltage by the simple relationship: \[ KE = eV \] where \( KE \) is the kinetic energy, \( e \) is the elementary charge, and \( V \) is the voltage. This makes kinetic energy directly proportional to the accelerating voltage.

• As voltage increases, so does the kinetic energy of electrons.
• Higher kinetic energy leads to shorter wavelengths and higher resolution images.

Planck's Constant

Planck's constant is a fundamental constant in physics, symbolized by \( h \). It is essential in the field of quantum mechanics, being a core component of the formula for the de Broglie wavelength.
Planck's constant allows us to calculate a particle's wavelength from its momentum. In the context of electron microscopes, this is crucial for determining how we can influence the wavelength through momentum changes driven by kinetic energy and voltage.

• Planck's constant \( (h) \) is approximately \( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \).
• A key part of de Broglie's equation: \( \lambda = \frac{h}{p} \).
• It helps in understanding the wave-particle duality of matter.

Understanding Planck's constant's role in the equations helps explain how we manipulate and interpret the behavior of electrons and other particles in electron microscopes.