Question

The potential difference to accelerate an electron was quadrupled. By what factor does the de Broglie wavelength of the electron beam change?

Answer 1

Correct Answer: 0.5

To determine the factor by which the de Broglie wavelength of an electron changes when the potential difference used to accelerate it is quadrupled, we start with the relevant formulae. The de Broglie wavelength λ is given by:

$$ \lambda = \frac{h}{p} $$

where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

The momentum \( p \) can be expressed in terms of the kinetic energy \( K \) as:

$$ p = \sqrt{2mK} $$

The kinetic energy \( K \) acquired by the electron is due to the potential difference \( V \), so \( K = eV \), where \( e \) is the charge of the electron.

Thus, the momentum becomes:

$$ p = \sqrt{2meV} $$

Substitute this expression for \( p \) into the de Broglie wavelength formula:

$$ \lambda = \frac{h}{\sqrt{2meV}} $$

If the potential difference \( V \) is quadrupled, the new wavelength \( \lambda' \) when the potential is \( 4V \) is:

$$ \lambda' = \frac{h}{\sqrt{2m \cdot 4eV}} $$

Simplifying the expression gives:

$$ \lambda' = \frac{h}{2\sqrt{2meV}} $$

Now, compare \( \lambda' \) with the original \( \lambda \):

$$ \frac{\lambda'}{\lambda} = \frac{\frac{h}{2\sqrt{2meV}}}{\frac{h}{\sqrt{2meV}}} = \frac{1}{2} $$

This means the de Broglie wavelength is halved. The factor by which the wavelength changes is 0.5.