Question

An electron is accelerated through an electric potential to a kinetic energy of \(13.4 \mathrm{keV}\). What is its characteristic wavelength? [Hint: Recall that the kinetic energy of a moving object is \(E=\frac{1}{2} m v^{2}\), where \(m\) is the mass of the object and \(\nu\) is the speed of the object.]

Short answer

The characteristic wavelength of the electron accelerated through the electric potential is approximately \(3.318 \times 10^{-11} \mathrm{m}\).

Step-by-step solution

Convert the given kinetic energy to Joules

We're given the electron's kinetic energy as 13.4 keV. Since 1 keV is equal to \(1.602 \times 10^{-16} \mathrm{J}\), let's first convert this energy into Joules: \(E = 13.4 \mathrm{keV} \times \frac{1.602 \times 10^{-16} \mathrm{J}}{1 \mathrm{keV}} = 2.148 \times 10^{-15} \mathrm{J}\)

Use the Kinetic Energy formula to find the electron's speed

We'll use the Kinetic Energy formula that is given by \(E = \frac{1}{2} m v^{2}\) where \(E\) is the energy of the electron, \(m\) is its mass, and \(v\) is its velocity. The mass of an electron is \(m = 9.109 \times 10^{-31} \mathrm{kg}\). Solve the equation for the velocity "v": \(v = \sqrt{\frac{2E}{m}}\) Substitute the given values for E and m: \(v = \sqrt{\frac{2 \times 2.148 \times 10^{-15} \mathrm{J}}{9.109 \times 10^{-31} \mathrm{kg}}}\)

Calculate the electron's velocity

Plug the numbers into a calculator and find the value of 'v': \(v = 2.197 \times 10^6 \mathrm{m/s}\)

Use the de Broglie wavelength formula

Now that we have the velocity, we can use the de Broglie wavelength formula to find the characteristic wavelength of the electron. The formula is given by: \(\lambda = \frac{h}{p} = \frac{h}{m v}\) Here, \(\lambda\) is the wavelength, \(h\) is the Planck's constant (\(h = 6.626 \times 10^{-34} \mathrm{J \cdot s}\)), \(m\) is the mass of the electron, and \(v\) is its velocity. Substitute the values we've found for m and v into the formula: \(\lambda = \frac{6.626 \times 10^{-34} \mathrm{J \cdot s}}{9.109 \times 10^{-31} \mathrm{kg} \times 2.197 \times 10^6 \mathrm{m/s}}\)

Solve for the characteristic wavelength

Calculate the value of the characteristic wavelength: \(\lambda = 3.318 \times 10^{-11} \mathrm{m}\) The characteristic wavelength of the electron accelerated through the electric potential is approximately \(3.318 \times 10^{-11} \mathrm{m}\).

Key concepts

Electron Velocity

When an electron is accelerated through an electric potential, it gains speed, creating velocity. Understanding this velocity is crucial for calculating other properties, such as the electron's wavelength using the de Broglie equation. The velocity of an electron can be figured out through its kinetic energy, which is the energy it possesses due to motion.

To find the velocity, you can use the kinetic energy formula:

• Given by: \( E = \frac{1}{2} m v^{2} \)
• Where \( E \) is the kinetic energy, \( m \) is the mass of the electron, and \( v \) is its velocity.

From the formula, we can rearrange to solve for velocity:
\[ v = \sqrt{\frac{2E}{m}} \] This formula allows us to plug in the known kinetic energy and mass of the electron to find its velocity. It shows the direct relationship between energy and velocity. Higher kinetic energy results in a higher velocity. For our example, the velocity was calculated as \( 2.197 \times 10^6 \ \mathrm{m/s} \).

Knowing electron velocity is crucial for determining its wavelength using the de Broglie equation.

Kinetic Energy Conversion

Kinetic energy, a form of energy an object possesses due to its motion, plays a pivotal role in understanding the behavior of electrons. In our problem, the given kinetic energy of the electron is expressed in kilo-electronvolts (keV), a common unit in atomic physics.

To use it in equations, we need to convert it from keV to the more universal unit of Joules (J). The conversion factor is:

• 1 keV = \( 1.602 \times 10^{-16} \ \mathrm{J} \)

Hence, the conversion for our given energy of 13.4 keV becomes:
\[ E = 13.4 \ \mathrm{keV} \times \frac{1.602 \times 10^{-16} \mathrm{J}}{1 \mathrm{keV}} = 2.148 \times 10^{-15} \ \mathrm{J} \]
Understanding and calculating accurate conversions is important because most equations in physics, particularly those involving energy and quantum mechanics, require energy in Joules. This consistency allows us to seamlessly apply converted values into different formulas, such as calculating velocity or de Broglie wavelength.

Planck's Constant

Planck's constant, denoted as \( h \), is a fundamental constant crucial to quantum mechanics. It relates the energy of photons to their frequency and also ties into the particle-wave duality through de Broglie's equation:

• \( h = 6.626 \times 10^{-34} \ \mathrm{J \cdot s} \)

In de Broglie's theory, any moving particle or object has an associated wave, and Planck's constant helps describe its wavelength:
\[ \lambda = \frac{h}{m v} \]In this formula, \( \lambda \) represents the wavelength, \( m \) is the mass, and \( v \) is the velocity. The denominator \( m v \) is the momentum, commonly seen as \( p \), so:
\[ \lambda = \frac{h}{p} \]
Planck's constant plays a pivotal role in understanding the particle-wave nature of electrons. It allows us to compute the standard wavelength of electrons when they move, which is instrumental in various applications like electron microscopy and quantum computing. Understanding Planck's constant deepens our grasp of how microscopic entities behave differently from larger, macroscopic systems.