Question

Which one of the following is correct with respect to an electron and a proton having the same de-Broglie wavelength of \(0.2\ \text{\AA}\)?

• A. Both have the same kinetic energy
• B. Both have the same velocity
• C. Both have the same momentum
• D. The kinetic energy of the proton is more than that of the electron

Hint:

de-Broglie wavelength depends only on momentum: same \(\lambda\) means same \(p\).

Answer 1

The Correct Option is C

Step 1: The de-Broglie wavelength relates to momentum by \(\lambda = \dfrac{h}{p}\), so \(p = \dfrac{h}{\lambda}\).

Step 2: Both particles have the same wavelength \(\lambda = 0.2\ \text{\AA}\), and \(h\) is a universal constant. Therefore both have identical momentum: \[p_e = p_p = \frac{h}{\lambda}.\]

Step 3: Check velocity: \(v = p/m\). Since \(m_p \gg m_e\) but \(p\) is equal, the electron moves much faster than the proton, so velocities differ. Option (B) is wrong.

Step 4: Check kinetic energy: \(K = \dfrac{p^2}{2m}\). With equal \(p\), \(K \propto 1/m\), so the lighter electron has the larger kinetic energy. This rules out (A) and (D).

Step 5: Only the momentum is common to both particles. \[\boxed{p_e = p_p = \frac{h}{\lambda}}\]