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

Concept: According to the de Broglie hypothesis, particles such as electrons exhibit wave-like behavior. The wavelength associated with a moving particle is given by \[ \lambda = \frac{h}{p} \] where

• \(h\) = Planck’s constant
• \(p\) = Momentum of the particle

Step 1: Determine the kinetic energy of the electron.} When an electron is accelerated through a potential difference \(V\), \[ \text{K.E.} = eV \] where \(e\) is the charge of the electron.
Step 2: Relate kinetic energy and momentum.} \[ \text{K.E.} = \frac{p^2}{2m} \] Thus, \[ eV = \frac{p^2}{2m} \]
Step 3: Solve for momentum.} \[ p = \sqrt{2meV} \]
Step 4: Substitute into de Broglie equation.} \[ \lambda = \frac{h}{p} \] \[ \lambda = \frac{h}{\sqrt{2meV}} \] Thus, the de Broglie wavelength is \[ \lambda = \frac{h}{\sqrt{2meV}} \]