Question

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of \(100V\).

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

Hint:

For electrons accelerated through a potential \(V\), remember the quick formula \[ \lambda = \frac{12.27}{\sqrt{V}} \ \text{\AA} \] This shortcut avoids lengthy calculations using \( \lambda = \frac{h}{p} \).

Answer 1

The Correct Option is B

Topic: Dual Nature of Matter and Radiation (Modern Physics)
Step 1: Understanding the Question:
We need to find the de Broglie wavelength (\(\lambda\)) associated with an electron that has been accelerated by an electric potential \(V\).
Step 2: Key Formula or Approach:
The de Broglie wavelength is given by \(\lambda = \frac{h}{p}\).
For an electron accelerated through a potential \(V\), the kinetic energy is \(eV\).
The simplified formula for electrons is:
\[ \lambda = \frac{12.27}{\sqrt{V}} \ \text{\AA} \]
Step 3: Detailed Explanation:
Given potential difference, \(V = 100 \text{ V}\).
Using the formula:
\[ \lambda = \frac{12.27}{\sqrt{100}} \]
Calculating the square root:
\[ \sqrt{100} = 10 \]
\[ \lambda = \frac{12.27}{10} = 1.227 \ \text{\AA} \]
Step 4: Final Answer:
The de Broglie wavelength of the electron is \(1.227 \ \text{\AA}\).