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A proton starts from rest and is accelerated through a potential difference V, acquiring a de Broglie wavelength lambda. Through what potential difference must an alpha particle be accelerated from rest so that it acquires the same de Broglie wavelength lambda?
- V volt
- 4V volt
- 2V volt
- V/8 volt
Correct answer: V/8 volt
Solution
De Broglie wavelength lambda = h / sqrt(2mqV). For equal wavelengths, mₚ * e * V = 4mₚ * 2e * V', giving V' = V/8.
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