Question

A stream of electrons from a heated filament was passed between two charged plates at a potential difference \( V \) volt. If \( e \) and \( m \) are the charge and mass of an electron, then the value of \( \frac{h}{\lambda} \) is:

• A. \( \sqrt{meV} \)
• B. \( \sqrt{2meV} \)
• C. \( meV \)
• D. \( 2meV \)

Hint:

Electrons gain kinetic energy when accelerated through a potential difference, affecting their de-Broglie wavelength.

Answer 1

The Correct Option is B

Step 1: {Calculate Electron Kinetic Energy}
\[KE = eV\]Step 2: {Apply de-Broglie Wavelength Formula}
\[\lambda = \frac{h}{\sqrt{2mKE}}\]Step 3: {Isolate \( \frac{h}{\lambda} \)}
\[\frac{h}{\lambda} = \sqrt{2meV}\]Consequently, option (B) is the correct solution.