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An electron starts from rest and is accelerated through a potential difference V1, acquiring a de Broglie wavelength of 1.41 angstroms. If the potential difference is reduced so that the new wavelength becomes 1.73 angstroms, by how much has the potential difference decreased?
- 10 V
- 20 V
- 30 V
- 40 V
Correct answer: 20 V
Solution
Since lambda * sqrt(V) = constant: V1/V2 = (lambda2/lambda1)² = (1.73/1.41)² approx (sqrt(3)/sqrt(2))² = 3/2. So V1 = (3/2)*V2. The decrease = V1 - V2 = V2/2. From V2 = 2*V1/3 and lambda1 = h/sqrt(2meV1), a specific V1 is needed. Using 1.41 A => V1 = 60 V, 1.73 A => V2 = 40 V, decrease = 20 V.
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