Question

The de-Broglie wavelength of an electron accelerated from rest through a potential difference of $100\text{ V}$ is approximately:

• A. \( 1.227 \text{ \AA} \)
• B. \( 12.27 \text{ \AA} \)
• C. \( 0.1227 \text{ \AA} \)
• D. \( 122.7 \text{ \AA} \)

Hint:

To absolute guarantee you don't misplace a decimal place during choice matching, remember that \(\sqrt{100}\) scales exactly by a factor of 10 in the denominator. This shifts the decimal point of your numerator constant (\(12.27\)) exactly one spot to the left, pointing directly to \(1.227\text{ \AA}\).

Answer 1

The Correct Option is A

Concept: According to the de-Broglie hypothesis, a moving material particle behaves as a wave, and its dual matter wavelength (\(\lambda\)) is inversely proportional to its linear momentum (\(p\)): \[ \lambda = \frac{h}{p} \] When a particle of charge \(q\) and mass \(m\) is accelerated from rest through an electric potential difference \(V\), the work done by the electric field transforms entirely into its translational kinetic energy (\(K\)): \[ K = qV \] Since kinetic energy is related to momentum by \(p = \sqrt{2mK}\), substituting this relation into the de-Broglie equation gives: \[ \lambda = \frac{h}{\sqrt{2mqV}} \]

Step 1: Applying the constants for an electron to simplify the radical expression.
For an electron, we substitute the standard known physical constants:
• Planck's constant, \(h \approx 6.626 \times 10^{-34} \text{ J s}\)
• Mass of an electron, \(m \approx 9.1 \times 10^{-31} \text{ kg}\)
• Charge of an electron, \(q = e \approx 1.6 \times 10^{-19} \text{ C}\) Substituting these constant properties collapses the fixed variables into a highly convenient, condensed shortcut formula specifically calibrated for electrons: \[ \lambda_{\text{electron}} = \frac{12.27}{\sqrt{V}} \text{ \AA} \quad \left(\text{where } 1 \text{ \AA} = 10^{-10}\text{ m}\right) \]

Step 2: Calculating the wavelength for $V = 100\text{V}$.
Given the accelerating potential difference \(V = 100\text{V}\). Substitute this value into our shortcut relation: \[ \lambda_{\text{electron}} = \frac{12.27}{\sqrt{100}} \text{ \AA} \] Since \(\sqrt{100} = 10\): \[ \lambda_{\text{electron}} = \frac{12.27}{10} = 1.227 \text{ \AA} \]