Question:medium

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

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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.
Updated On: Apr 20, 2026
  • 1.227 nm
  • 0.613 nm
  • 0.302 nm
  • 0.153 nm
  • 2.454 nm
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
According to de Broglie's hypothesis, a moving particle like an electron has a wave nature with an associated wavelength.
When an electron is accelerated from rest through a potential difference, it gains kinetic energy, which can be directly related to its de Broglie wavelength.
Step 2: Key Formula or Approach:
The specific formula for the de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ (in volts) is given by $\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}$.
Step 3: Detailed Explanation:
We are given the accelerating potential $V = 64 \text{ V}$.
Substitute this value directly into the empirical formula:
\[ \lambda = \frac{1.227}{\sqrt{64}} \text{ nm} \]
Calculate the square root of the denominator:
\[ \lambda = \frac{1.227}{8} \text{ nm} \]
Perform the final division:
\[ \lambda = 0.153375 \text{ nm} \]
Rounding to three decimal places to match the given options gives $0.153 \text{ nm}$.
Step 4: Final Answer:
The de Broglie wavelength associated with the electron is 0.153 nm.
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