Question

The de Broglie wavelength associated with an electron accelerated by a potential of 64 V is ____.

• A. 1.227 nm
• B. 0.613 nm
• C. 0.302 nm
• D. 0.153 nm
• E. 2.454 nm

Hint:

Notice that $64$ is a perfect square. In multiple-choice questions, potential values are often chosen (like 100, 64, or 25) so that the square root is an integer, making the calculation much faster.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
According to de Broglie's hypothesis, moving particles like electrons exhibit wave-like properties. The wavelength depends on the momentum of the electron, which is gained through acceleration in an electric potential.

Step 2: Key Formula or Approach
For an electron accelerated through a potential $V$, the shortcut formula for the wavelength in nanometers (nm) is: \[ \lambda = \frac{1.227}{\sqrt{V}} \, \text{nm} \]

Step 3: Detailed Explanation
1. Identify the accelerating potential: $V = 64 \, \text{V}$.
2. Substitute into the formula: \[ \lambda = \frac{1.227}{\sqrt{64}} \]
3. Calculate the square root: $\sqrt{64} = 8$.
4. Perform the division: \[ \lambda = \frac{1.227}{8} = 0.153375 \, \text{nm} \]

Step 4: Final Answer
The de Broglie wavelength is approximately 0.153 nm.