Step 1: Understanding the Question:
We need to find the de-Broglie wavelength associated with an electron after it has been accelerated by a given potential difference.
Step 2: Key Formula or Approach:
The de-Broglie wavelength \( \lambda \) is given by \( \lambda = h/p \), where \( h \) is Planck's constant and \( p \) is momentum.
The kinetic energy gained by an electron accelerated through a potential \( V \) is \( K = eV \).
Since \( K = p^2/(2m) \), we have \( p = \sqrt{2mK} = \sqrt{2meV} \).
This gives the general formula: \( \lambda = \frac{h}{\sqrt{2meV}} \).
For an electron, a convenient shortcut formula exists:
\[ \lambda (\text{in Ångströms}) \approx \frac{12.27}{\sqrt{V (\text{in Volts})}} \]
Step 3: Detailed Explanation:
Given potential difference \( V = 100 \) V.
Using the shortcut formula:
\[ \lambda = \frac{12.27}{\sqrt{100}} \, \text{Å} \]
\[ \lambda = \frac{12.27}{10} \, \text{Å} \]
\[ \lambda = 1.227 \, \text{Å} \]
Note on options:
Option (A) is \(1.227\ \text{Å}\).
Option (B) is \(0.1227\ \text{nm}\). Since \(1 \text{ nm} = 10 \text{ Å}\), \(1.227 \text{ Å}\) is equal to \(0.1227 \text{ nm}\). Both options (A) and (B) are numerically correct representations of the same length. In such cases, one must choose the most direct answer or what is conventionally used. The formula directly gives the answer in Ångströms.
Step 4: Final Answer:
The de-Broglie wavelength is \( 1.227 \, \text{Å} \).