Question

An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is $1.227 \times 10^{-2}nm$, the potential difference is :

• A. $10$ V
• B. $10^2$ V
• C. $10^3$ V
• D. $10^4$ V

Answer 1

The Correct Option is D

To find the potential difference through which an electron is accelerated, we need to use the de Broglie wavelength formula and the relationship between kinetic energy and potential difference.

The de Broglie wavelength \((\lambda)\) of a particle like an electron is given by:

\(\lambda = \frac{h}{p}\)

where \(h\) is Planck's constant and \(p\) is the momentum of the particle. However, the momentum of an electron that has been accelerated from rest by a potential difference \(V\) is given by:

\(p = \sqrt{2m e V}\)

where \(m\) is the mass of the electron and \(e\) is the charge of the electron.

Substituting the expression for \(p\) in the de Broglie formula, we have:

\(\lambda = \frac{h}{\sqrt{2m e V}}\)

Rearranging for \(V\), we get:

\(V = \frac{h^2}{2m e \lambda^2}\)

Given: \(\lambda = 1.227 \times 10^{-2} \text{ nm} = 1.227 \times 10^{-11} \text{ m}\)

Planck's constant, \(h = 6.626 \times 10^{-34} \text{ Js}\)

Electron mass, \(m = 9.109 \times 10^{-31} \text{ kg}\)

Electron charge, \(e = 1.602 \times 10^{-19} \text{ C}\)

Substituting these values into the formula for \(V\):

\(V = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times (1.227 \times 10^{-11})^2}\)

Calculating this gives:

\(V \approx 10^4 \text{ V}\)

Thus, the potential difference is \(10^4\) V. This matches the correct answer provided.

Hence, the correct option is \(10^4\) V.