Question

What is the de Broglie wavelength of an electron accelerated through a potential difference of \(100\,\text{V}\)?

• 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 \(V\), quickly use \[ \lambda(\text{\AA}) = \frac{12.27}{\sqrt{V}} \] This shortcut is widely used in quantum mechanics problems.

Answer 1

The Correct Option is B

Topic - Dual Nature of Matter (de Broglie Wavelength):
This topic explores the wave nature of particles. For charged particles like electrons, the wavelength depends on the accelerating potential.
Step 1: Understanding the Question:
The objective is to calculate the de Broglie wavelength for an electron given its accelerating voltage \(V = 100\,\text{V}\).
Step 2: Key Formula or Approach:
For an electron accelerated from rest through a potential \(V\), the wavelength \(\lambda\) is: \[ \lambda = \frac{h}{\sqrt{2meV}} \] In a simplified numerical form for electrons: \[ \lambda = \frac{12.27}{\sqrt{V}} \ \text{\AA} \] Step 3: Detailed Solution:
1. Identify the given value: \(V = 100\,\text{V}\).
2. Substitute the value into the simplified formula:
\[ \lambda = \frac{12.27}{\sqrt{100}} \ \text{\AA} \] 3. Since \(\sqrt{100} = 10\):
\[ \lambda = \frac{12.27}{10} \ \text{\AA} \] \[ \lambda = 1.227 \ \text{\AA} \] Step 4: Final Answer:
The de Broglie wavelength is \(1.227 \ \text{\AA}\).