Question

What is the de Broglie wavelength of an electron accelerated through a potential difference of \(V\) volts?

• A. \(\dfrac{h}{mv}\)
• B. \(\dfrac{h}{\sqrt{2meV}}\)
• C. \(\dfrac{h}{eV}\)
• D. \(\sqrt{\dfrac{2eV}{m}}\)

Hint:

For an electron accelerated through potential \(V\): \[ \lambda = \frac{h}{\sqrt{2meV}} \] or approximately \[ \lambda(\text{Å}) = \frac{12.27}{\sqrt{V}} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Every moving particle behaves like a wave, called matter waves. The wavelength associated with this wave is the de Broglie wavelength, which depends on the particle's momentum.
Step 2: Key Formula or Approach:
The fundamental de Broglie wavelength formula is:
\[ \lambda = \frac{h}{p} \]
The kinetic energy $K$ gained by an electron of charge $e$ accelerated through a potential difference $V$ is $K = eV$.
The relation between momentum $p$ and kinetic energy $K$ is $p = \sqrt{2mK}$.
Step 3: Detailed Explanation:
First, find the momentum of the electron using its kinetic energy:
\[ p = \sqrt{2m(eV)} = \sqrt{2meV} \]
Next, substitute this momentum back into the de Broglie wavelength equation:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]
This matches the expression in option (B).
Step 4: Final Answer:
The correct option is (B).