Question

A proton accelerated through a potential difference of 100 V has de-Broglie wavelength \( \lambda_0 \). The de-Broglie wavelength of an \(\alpha\)-particle accelerated through 800 V is:

• A. \( \frac{\lambda_0}{\sqrt{2}} \)
• B. \( \frac{\lambda_0}{2} \)
• C. \( \frac{\lambda_0}{4} \)
• D. \( \frac{\lambda_0}{8} \)

Hint:

For charged particles: \( \lambda \propto \frac{1}{\sqrt{mqV}} \) — memorize this for quick comparisons!

Answer 1

The Correct Option is D

Concept: De-Broglie wavelength: \[ \lambda = \frac{h}{\sqrt{2mqV}} \] \[ \lambda \propto \frac{1}{\sqrt{mqV}} \]

Step 1: Proton case \[ \lambda_0 \propto \frac{1}{\sqrt{m \cdot e \cdot 100}} \]

Step 2: Alpha particle \[ m_\alpha = 4m,\quad q_\alpha = 2e,\quad V = 800 \] \[ \lambda_\alpha \propto \frac{1}{\sqrt{4m \cdot 2e \cdot 800}} \]

Step 3: Ratio \[ \frac{\lambda_\alpha}{\lambda_0} = \sqrt{\frac{m \cdot e \cdot 100}{4m \cdot 2e \cdot 800}} \] \[ = \sqrt{\frac{100}{6400}} = \sqrt{\frac{1}{64}} = \frac{1}{8} \] \[ \therefore \lambda_\alpha = \frac{\lambda_0}{8} \]