Question

An electron is accelerated through an electric potential to a kinetic energy of \(2.15 \times 10^{-15} \mathrm{J}\) . 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 is approximately \(3.34 \times 10^{-10}\) meters.

Step-by-step solution

We are given the kinetic energy of the electron: \(E = 2.15 \times 10^{-15} J\). Using the formula for kinetic energy, \(E = \frac{1}{2}mv^2\), we can solve for the speed of the electron (v). Rearrange the formula to isolate v: \(v^2 = \frac{2E}{m}\) Note that the mass of an electron is approximately \(9.11 \times 10^{-31} kg\). Now, we substitute the given values and find the square root to find the speed of the electron: \begin{aligned} v &= \sqrt{\frac{2(2.15 \times 10^{-15})}{9.11 \times 10^{-31}}} \\ v &= \sqrt{4.72 \times 10^{14}} \\ v &= 2.17 \times 10^7 m/s \end{aligned} #Step 2: Find the characteristic wavelength#

Now that we have the speed of the electron, we can use the de Broglie wavelength formula \(\lambda = \frac{h}{mv}\) to find the characteristic wavelength. The Planck constant (h) has a value of \(6.63 \times 10^{-34} Js\). Plug in the values of mass, speed, and the Planck constant: \(\lambda = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(2.17 \times 10^7)}\) Calculate the wavelength: \(\lambda = 3.34 \times 10^{-10} m\) The characteristic wavelength of the electron is approximately \(3.34 \times 10^{-10}\) meters.

Key concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. It's a fundamental concept in physics that helps us understand how objects move and interact. You can calculate kinetic energy using the formula \(E = \frac{1}{2}mv^2\), where \(E\) is the kinetic energy, \(m\) is the mass of the object, and \(v\) is its velocity.
For an electron, knowing its kinetic energy is crucial to determine other properties like its speed. By rearranging the formula, you can solve for the velocity by finding that \(v = \sqrt{\frac{2E}{m}}\).
This approach makes it possible to calculate other physical characteristics, such as the de Broglie wavelength, which is essential in understanding quantum behavior.

Electron Mass

The mass of an electron is a fundamental constant in physics. It's very tiny—about \(9.11 \times 10^{-31} \text{ kg}\). Given its small size, the mass of an electron is significant in calculations involving atomic and subatomic particles.
Using accurate values for the electron mass is crucial in equations like kinetic energy and de Broglie wavelength. These calculations allow us to predict and explain the behavior of electrons in different contexts.
For example, when determining the speed or wavelength of an electron accelerated by a known energy, precise knowledge of its mass is required to obtain correct results. Such calculations underpin key discoveries and applications in quantum mechanics.

Planck Constant

The Planck constant \(h\) is a crucial part of quantum mechanics. It connects the energy of a photon to its frequency via the equation \(E = hf\), where \(f\) is the frequency of the photon.
In the context of de Broglie wavelength, the Planck constant helps relate particle-like and wave-like properties of subatomic particles through \(\lambda = \frac{h}{mv}\). Here, \(\lambda\) is the wavelength, \(m\) is mass, and \(v\) is velocity.
The constant \(h\) is a tiny value, approximately \(6.63 \times 10^{-34} \text{ Js}\). Its smallness corresponds to the equally minuscule scales at which quantum effects become significant, allowing us to explore and understand phenomena that are inaccessible at larger scales.
The Planck constant thus serves as a bridge between the classical and quantum worlds, playing an essential role in calculations of energy and motion at the microscopic level.