An electron is accelerated through an electric potential to a kin...

Question

An electron is accelerated through an electric potential to a kinetic energy of 1.6 * 10^-15 J. What is its characteristic wavelength? [Hint: Recall that the kinetic energy of a moving object is E = 1/2 mv^2, where m is the mass of the object and v is the speed of the object.]

Accepted Answer

Step 1: Identify the relevant equations. The de Broglie wavelength equation is $ \lambda = \frac{h}{p} $, where $ \lambda $ is the wavelength, $ h $ is Planck's constant, and $ p $ is the momentum of the electron. The momentum $ p $ can be expressed as $ p = mv $, where $ m $ is the mass and $ v $ is the velocity of the electron.. Step 2: Relate kinetic energy to velocity. The kinetic energy $ E $ of the electron is given by $ E = \frac{1}{2}mv^2 $. Rearrange this equation to solve for $ v $: $ v = \sqrt{\frac{2E}{m}} $.. Step 3: Calculate the momentum $ p $ using the velocity. Substitute the expression for $ v $ from Step 2 into the momentum equation $ p = mv $ to get $ p = m \sqrt{\frac{2E}{m}} $. Simplify this to $ p = \sqrt{2mE} $.. Step 4: Substitute the expression for momentum $ p $ into the de Broglie wavelength equation. This gives $ \lambda = \frac{h}{\sqrt{2mE}} $.. Step 5: Substitute known values into the equation. Use Planck's constant $ h = 6.626 \times 10^{-34} \text{ J s} $, the mass of an electron $ m = 9.109 \times 10^{-31} \text{ kg} $, and the given kinetic energy $ E = 1.6 \times 10^{-15} \text{ J} $ to calculate $ \lambda $.

An electron is accelerated through an electric potential to a kinetic energy of 1.6 * 10^-15 J. What is its characteristic wavelength? [Hint: Recall that the kinetic energy of a moving object is E = 1/2 mv^2, where m is the mass of the object and v is the speed of the object.]

Key Concepts

Kinetic Energy

De Broglie Wavelength

Planck's Constant