Question

Determine the de Broglie wavelength of an electron accelerated through \(100\,V\).

• A. \(0.1227\,\text{\AA}\)
• B. \(1.227\,\text{\AA}\)
• C. \(12.27\,\text{\AA}\)
• D. \(0.01227\,\text{\AA}\)

Hint:

Start from the energy gained by an electron, $K=eV$, and use it in $p=\sqrt{2m_eK}$. Once the momentum is known, apply $\lambda=h/p$ and convert metres to angstroms.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the matter wave (de Broglie wavelength) associated with an electron that has been accelerated from rest through a given electrostatic potential difference.
Step 2: Key Formula or Approach:
The general de Broglie wavelength formula is \(\lambda = \frac{h}{p}\).
For an electron specifically, when accelerated through a potential difference \(V\), the formula simplifies to a highly useful standard shortcut:
\[ \lambda = \frac{12.27}{\sqrt{V}} \, \text{\AA} \] where \(V\) is measured in volts.
Step 3: Detailed Solution:
We are given the accelerating potential:
\[ V = 100\,V \] Substitute \(V = 100\) directly into the specific electron formula:
\[ \lambda = \frac{12.27}{\sqrt{100}} \, \text{\AA} \] Since the square root of \(100\) is \(10\):
\[ \lambda = \frac{12.27}{10} \, \text{\AA} \] Divide by \(10\) by shifting the decimal point one place to the left:
\[ \lambda = 1.227 \, \text{\AA} \] Step 4: Final Answer:
The de Broglie wavelength of the electron is \(1.227\,\text{\AA}\).