Question:

The wavelength of a fast moving particle (\(\lambda\)) is related to its momentum (p) and Planck’s constant (h) as

Show Hint

Remember that high momentum (larger mass or higher velocity) means a shorter wavelength.
This is why macroscopic objects do not show noticeable wave properties.
Updated On: Apr 29, 2026
  • \(\lambda = hp\)
  • \(\lambda = h^2 / p\)
  • \(\lambda = h / p\)
  • \(\lambda = p^2 / h\)
  • \(\lambda = h / p^2\)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
According to de Broglie's hypothesis, every moving particle has an associated wave character.
The wavelength of this matter wave is called the de Broglie wavelength.

Step 3: Detailed Explanation:

The relationship established by Louis de Broglie relates the wavelength (\(\lambda\)) of a particle to its momentum (\(p\)) and Planck's constant (\(h\)).
The formula is:
\[ \lambda = \frac{h}{p} \]
Since momentum is the product of mass (\(m\)) and velocity (\(v\)), it can also be written as:
\[ \lambda = \frac{h}{mv} \]
This equation shows that the wavelength is inversely proportional to the momentum.

Step 4: Final Answer:

The correct mathematical relation is \(\lambda = h / p\).
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