Question

Determine the de Broglie wavelength of an electron (mass \(=9.11 \times 10^{-28} \mathrm{~g}\) ) having a kinetic energy of \(100 \mathrm{eV}\)

Short answer

The de Broglie wavelength of an electron with a mass of \(9.11 \times 10^{-31}\mathrm{~kg}\) and a kinetic energy of \(100 \mathrm{eV}\) is approximately \(1.227 \times 10^{-10} \mathrm{m}\).

Step-by-step solution

Convert the kinetic energy to Joules

Given the kinetic energy in electronvolts (eV), we need to convert it to Joules (J). We can do this using the conversion factor 1 eV = \(1.602 \times 10^{-19}\) J. \(100 \mathrm{eV} \times \frac{1.602 \times 10^{-19} \mathrm{J}}{1 \mathrm{eV}} = 1.602 \times 10^{-17} \mathrm{J}\)

Calculate the velocity of the electron using the kinetic energy

Now that we have the kinetic energy in Joules, we can use the kinetic energy equation to solve for the velocity of the electron. The kinetic energy equation is: \(KE = \frac{1}{2}mv^2\) where \(KE\) is the kinetic energy, \(m\) is the mass of the electron, and \(v\) is the velocity of the electron. We can rearrange the equation to solve for \(v\): \(v = \sqrt{\frac{2 \times KE}{m}}\) Plug in the converted kinetic energy and mass of the electron into the equation: \(v = \sqrt{\frac{2 \times (1.602 \times 10^{-17} \mathrm{J})}{(9.11 \times 10^{-31} \mathrm{~kg})}} \approx 5.932 \times 10^{5} \mathrm{m/s}\)

Calculate the momentum of the electron

Now that we have the velocity of the electron, we can use the equation for momentum to find the momentum of the electron: \(p = mv\) where \(p\) is the momentum, \(m\) is the mass of the electron, and \(v\) is the velocity of the electron. Plug in the mass and velocity values: \(p = (9.11 \times 10^{-31} \mathrm{~kg})(5.932 \times 10^{5} \mathrm{m/s}) \approx 5.404 \times 10^{-24} \mathrm{kg \cdot m/s}\)

Calculate the de Broglie wavelength of the electron

Finally, we can use the de Broglie wavelength formula to find the wavelength of the electron: \(\lambda = \frac{h}{p}\) where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), and \(p\) is the momentum of the electron. Plug in the values of Planck's constant and the momentum: \(\lambda = \frac{6.626 \times 10^{-34} \mathrm{Js}}{5.404 \times 10^{-24} \mathrm{kg \cdot m/s}} \approx 1.227 \times 10^{-10} \mathrm{m}\) So, the de Broglie wavelength of the electron is approximately \(1.227 \times 10^{-10} \mathrm{m}\).

Key concepts

Kinetic Energy

Kinetic energy is a key concept in physics, representing the energy that an object possesses due to its motion. For particles like electrons, kinetic energy can be expressed in electronvolts (eV), which is a common unit of energy in atomic and particle physics.

When calculating kinetic energy, especially in contexts like determining an electron's de Broglie wavelength, it's important to convert this energy into Joules (J) using the conversion factor: 1 eV = \(1.602 \times 10^{-19}\) J.

In the exercise, an electron has a kinetic energy of 100 eV. Multiplying this by the conversion factor gives us \(1.602 \times 10^{-17} \text{ J}\). This conversion allows us to use kinetic energy in calculations involving other physical quantities, like velocity.

Momentum

Momentum is another fundamental quantity in physics, defined as the product of an object's mass and velocity. It explains how much motion an object has and how difficult it is to stop it.

For the electron in our exercise, its momentum \( p \) is calculated using the formula \( p = mv \), where \( m \) is the mass of the electron and \( v \) is its velocity. For our case, \( p = (9.11 \times 10^{-31} \text{ kg})(5.932 \times 10^{5} \text{ m/s}) \approx 5.404 \times 10^{-24} \text{ kg m/s} \).

Understanding momentum helps us to connect an electron's motion to its wave-like properties, which are expressed through the de Broglie wavelength.

Planck's Constant

Planck's constant \( h \) is a significant and fundamental constant in quantum mechanics, linking the wavelength of particles to their momentum. Its value is \(6.626 \times 10^{-34} \text{ Js}\).

This constant plays a crucial role in the de Broglie hypothesis, where the de Broglie wavelength \( \lambda \) of a particle is given by \( \lambda = \frac{h}{p} \), with \( p \) being momentum.

Planck's constant essentially quantifies the size of quanta, the smallest possible units of energy, making it pivotal for relating particle properties to wave properties in quantum physics.

Electron Mass

The mass of an electron is a vital factor when working with particle physics equations such as kinetic energy and momentum. The mass of an electron is \(9.11 \times 10^{-31} \text{ kg}\).

This precise value is needed for accurate calculations in determining properties like velocity or momentum. For kinetic exercises, electron mass is used within energy equations, like \( KE = \frac{1}{2}mv^2 \), to resolve the velocity or further calculations for de Broglie wavelength.

The seemingly tiny mass signifies the low inertia of electrons, highlighting the significant effect even a small amount of external energy can have on their velocity and momentum.

Velocity Calculation

Velocity in physics is the speed and direction of an object's motion. Calculating an electron's velocity from its kinetic energy is crucial for determining its wave properties as described by de Broglie's theory.

From the kinetic energy equation \( KE = \frac{1}{2}mv^2 \), the velocity \( v \) can be found using: \( v = \sqrt{\frac{2 KE}{m}} \).

For our electron, using the kinetic energy \(1.602 \times 10^{-17}\) J and electron mass \(9.11 \times 10^{-31} \text{ kg}\), we find \( v \approx 5.932 \times 10^{5} \text{ m/s}\).

This fast velocity reflects how quickly the electron moves when subjected to high energy, illustrating abstract wave-particle duality concepts in quantum mechanics. Understanding velocity through kinetic equations lays the groundwork for exploring deeper quantum phenomena.