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: {Finding Electron Kinetic Energy}
\[ KE = eV \] Step 2: {Using de-Broglie Wavelength Formula}
\[ \lambda = \frac{h}{\sqrt{2mKE}} \] Step 3: {Rearranging for \( \frac{h}{\lambda} \)}
\[ \frac{h}{\lambda} = \sqrt{2meV} \] Thus, the correct answer is (B).

Answer 2

Step 1: Understand the physical setup
- Electrons are accelerated by a potential difference \( V \).
- This means each electron gains kinetic energy due to the electric field.

Step 2: Apply the energy conservation principle
The kinetic energy gained by an electron is equal to the electrical energy supplied:
\( \frac{1}{2}mv^2 = eV \)
Where:
- \( m \) = mass of electron
- \( v \) = final velocity of electron
- \( e \) = charge of electron
- \( V \) = accelerating voltage

Step 3: Use de Broglie relation
According to de Broglie's hypothesis:
\( \lambda = \frac{h}{p} = \frac{h}{mv} \)
So,
\( \frac{h}{\lambda} = mv \)

Step 4: Substitute \( mv \) using the energy equation
From \( \frac{1}{2}mv^2 = eV \), we get:
\( v = \sqrt{\frac{2eV}{m}} \)
Then,
\( mv = m \cdot \sqrt{\frac{2eV}{m}} = \sqrt{2meV} \)

Step 5: Final Answer
The value of \( \frac{h}{\lambda} \) is:
\( \sqrt{2meV} \)