Question

In an electron microscope, electrons are accelerated by passing them through a voltage difference. The kinetic energy thus acquired by the electrons is equal to the voltage times the charge on the electron. Thus a voltage difference of 1 volt imparts a kinetic energy of $1.602\times {10}^{-19}$ volt-coulomb or $1.602\times {10}^{-19}\mathrm{J}$. Calculate the wavelength associated with electrons accelerated by $5.00\times {10}^3$ volts.

Short answer

The wavelength is approximately $5.48\times {10}^{-12}\mathrm{m}$.

Step-by-step solution

Understand the Relationship Between Voltage and Kinetic Energy

The kinetic energy $KE$ of the electrons after being accelerated through a voltage $V$ is given by the equation $KE=eV$, where $e=1.602\times {10}^{-19}$ coulombs is the charge of an electron. For a voltage difference of $5.00\times {10}^3$ volts, the kinetic energy gained by the electron is $KE=5.00\times {10}^3\times 1.602\times {10}^{-19}\mathrm{J}$.

Calculate the Kinetic Energy

Substitute the values into the kinetic energy equation: $$KE=5.00\times {10}^3\times 1.602\times {10}^{-19}\mathrm{J}=8.01\times {10}^{-16}\mathrm{J}$$This is the kinetic energy of each electron after acceleration.

Relate Kinetic Energy to Momentum

The momentum $p$ of the electron can be related to its kinetic energy using the equation $KE=\frac{p^2}{2m}$, where $m$ is the mass of the electron ($9.11\times {10}^{-31}\mathrm{kg}$). Rearrange the equation to solve for $p$:$$p=\sqrt{2m\cdot KE}$$

Calculate Momentum of the Electron

Substitute the values into the momentum equation:$$p=\sqrt{2\times 9.11\times {10}^{-31}\mathrm{kg}\times 8.01\times {10}^{-16}\mathrm{J}}$$ $$p=\sqrt{1.46182\times {10}^{-45}}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}\approx 1.21\times {10}^{-22}\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$$

Use de Broglie Wavelength Formula

The de Broglie wavelength $\lambda$ of an electron is given by the equation $\lambda =\frac{h}{p}$, where $h=6.626\times {10}^{-34}{\mathrm{m}}^2\cdot \mathrm{kg}/\mathrm{s}$ is Planck's constant. Substitute the values:$$\lambda =\frac{6.626\times {10}^{-34}}{1.21\times {10}^{-22}}\mathrm{m}\approx 5.48\times {10}^{-12}\mathrm{m}$$

Conclude with the Wavelength Calculation

After performing the calculations, we find that the wavelength associated with electrons accelerated by $5.00\times {10}^3$ volts is approximately $5.48\times {10}^{-12}\mathrm{m}$. This is the wavelength of the electrons in an electron microscope at this voltage.

Key concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics, representing the energy that an object possesses due to its motion. In the context of an electron microscope, when electrons are accelerated, they gain kinetic energy as a result of moving through a voltage difference. The formula to calculate the kinetic energy, $KE$, of these electrons is given by:
$$KE=eV$$where $e=1.602\times {10}^{-19}$ coulombs is the charge of an electron, and $V$ is the voltage difference across which the electron is accelerated.

• The kinetic energy acquired is directly proportional to the voltage difference.
• For example, a voltage of 1 volt imparts to an electron a kinetic energy of $1.602\times {10}^{-19}$ joules.
• This formula is crucial in applications involving particle acceleration, like in electron microscopes.

Understanding how kinetic energy relates to voltage is essential for calculating other properties, such as the electron’s wavelength.

Voltage Difference

The concept of voltage difference, also known simply as voltage, is a measure of the electric potential difference between two points. It is the driving force that propels electrons in an electric circuit or in devices like an electron microscope.

• Voltage is measured in volts (V).
• It determines how much energy per unit charge is applied to electrons.
• In an electron microscope, a voltage difference causes electrons to accelerate and gain kinetic energy.

In this context, the higher the voltage, the greater the energy imparted to the electrons, which is then manifested in their kinetic energy. A voltage difference of $5.00\times {10}^3$ volts, for instance, is used to accelerate electrons in an electron microscope, providing them with a significant amount of kinetic energy to reveal very small structures.

de Broglie Wavelength

The de Broglie wavelength is a key concept in quantum mechanics that relates the wave-like behavior of particles to their momentum. The concept arises from the pioneering work of physicist Louis de Broglie, who proposed that particles like electrons exhibit both wave and particle characteristics. The de Broglie wavelength, $\lambda$, is calculated using:
$$\lambda =\frac{h}{p}$$where $h$ is Planck's constant, and $p$ is the momentum of the particle.

• The wavelength is inversely proportional to the electron's momentum.
• Electrons with high momentum have shorter wavelengths.
• This wavelength is a factor in electron microscopes, determining the resolution of the images produced.

For an electron accelerated by a voltage of $5.00\times {10}^3$ volts, the de Broglie wavelength can be computed. This calculation allows us to see the level of detail that electron microscopes can achieve.

Planck's Constant

Planck's constant, denoted as $h$, is a fundamental constant in physics that connects the energy of photons to their frequency. Its value is $6.626\times {10}^{-34}{\mathrm{m}}^2\cdot \mathrm{kg}/\mathrm{s}$. This constant plays a critical role not only in quantum mechanics but also in determining the de Broglie wavelength of particles like electrons.

• Essential for calculations involving particle wavelengths.
• Used to relate momentum and wavelength in the de Broglie equation.
• Highlights the quantized nature of energy transfers at microscopic scales.

In the context of electron microscopes, using Planck's constant to calculate the de Broglie wavelength is vital in understanding the wave properties of electrons. This enables scientists and engineers to exploit these properties for achieving high-resolution imaging, crucial for examining tiny and complex structures.