Question

If the de Broglie wavelengths of proton p and alpha particle \(\alpha\) are same, then

• A. both have same momentum
• B. both have same energy
• C. momentum of p is twice that of \(\alpha\)
• D. momentum of \(\alpha\) is twice that of p
• E. energy of p is twice that of \(\alpha\)

Hint:

Wavelength is purely a function of momentum in de Broglie's theory.
If \(\lambda_1 = \lambda_2\), then \(p_1 = p_2\).
If you need to compare kinetic energy (\(K\)), use \(K = p^2 / 2m\). Since masses differ, their energies would be different even with the same momentum.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The de Broglie hypothesis states that every moving particle has an associated wave character, with a wavelength inversely proportional to its momentum.

Step 2: Key Formula or Approach:
The de Broglie wavelength (\(\lambda\)) is given by:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the linear momentum.

Step 3: Detailed Explanation:
Given that the wavelengths are the same:
\[ \lambda_p = \lambda_\alpha \]
Substituting the formula:
\[ \frac{h}{p_p} = \frac{h}{p_\alpha} \]
Canceling the constant \(h\) from both sides:
\[ p_p = p_\alpha \]
This means that regardless of their masses or velocities, if their de Broglie wavelengths are identical, their momenta must be identical.

Step 4: Final Answer:
Both have the same momentum.