De Broglie Wavelength of Electron Accelerated by Potential V

TITLE: De Broglie Wavelength of Electron Accelerated by Potential V DESCRIPTION: Learn how to calculate the de Broglie wavelength of an electron accelerated through a potential difference V using the formula λ = h/√(2meV).

Concept Overview

This question tests the understanding of wave-particle duality, specifically the de Broglie hypothesis which states that all matter exhibits wave-like properties. When a charged particle like an electron is accelerated through a potential difference, it gains kinetic energy. This kinetic energy can then be related to its momentum, which in turn determines its de Broglie wavelength. The core of this problem lies in connecting the electrical work done by the potential difference to the kinetic energy of the electron and then applying the de Broglie wavelength formula.

Step 1: Relate the work done by the potential difference to the kinetic energy gained by the electron. When an electron (charge $e$) is accelerated through a potential difference $V$, the work done on it by the electric field is converted into kinetic energy. The work done is given by the product of charge and potential difference. $W = qV$ For an electron, the charge is $e$, so the work done is $W = eV$. This work done equals the kinetic energy ($KE$) gained by the electron. $KE = eV$

Step 2: Express kinetic energy in terms of momentum. The kinetic energy of a particle is related to its momentum ($p$) by the formula: $KE = \frac{p^2}{2m}$ where $m$ is the mass of the particle. For an electron, $m = m_e$.

Step 3: Equate the two expressions for kinetic energy to find momentum. By equating the kinetic energy gained from the potential difference to the kinetic energy expressed in terms of momentum, we can find the momentum of the electron. $\frac{p^2}{2m_e} = eV$ Solving for $p^2$: $p^2 = 2m_e eV$ Taking the square root to find the momentum: $p = \sqrt{2m_e eV}$

Step 4: Apply the de Broglie wavelength formula. The de Broglie wavelength ($\lambda$) of a particle is given by the equation: $\lambda = \frac{h}{p}$ where $h$ is Planck's constant and $p$ is the momentum of the particle.

Step 5: Substitute the expression for momentum into the de Broglie wavelength formula. Now, substitute the expression for momentum ($p = \sqrt{2m_e eV}$) derived in Step 3 into the de Broglie wavelength formula from Step 4. $\lambda = \frac{h}{\sqrt{2m_e eV}}$ This is the de Broglie wavelength of an electron accelerated through a potential difference $V$.

Key Takeaways:

• Wave-particle duality is a fundamental concept where particles exhibit wave-like properties.
• The de Broglie wavelength relates a particle's momentum to its wavelength.
• The kinetic energy gained by a charged particle accelerated through a potential difference is equal to the work done by the electric field ($KE = qV$).
• The momentum of the electron can be found by equating its kinetic energy from the potential difference to the momentum-kinetic energy relationship ($p = \sqrt{2mKE}$).

Answer: $\boxed{\lambda = \frac{h}{\sqrt{2m_e eV}}}$