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A particle of mass m and charge q is accelerated through a potential difference of 50 V, producing a de Broglie wavelength lambda. If the mass is doubled (to 2m) while keeping the charge the same, and the new accelerating potential difference is 2500 V, what is the new de Broglie wavelength?
- 0.1 lambda
- sqrt(2)*lambda
- 10*lambda
- 10*sqrt(2)*lambda
Correct answer: 0.1 lambda
Solution
de Broglie wavelength: lambda = h/sqrt(2mqV). Original: lambda = h/sqrt(2*m*q*50). New: m' = 2m, V' = 2500 V. lambda' = h/sqrt(2*2m*q*2500) = h/sqrt(10000*m*q). Original denominator: h/sqrt(100*m*q). Ratio: lambda'/lambda = sqrt(100*m*q)/sqrt(10000*m*q) = sqrt(100/10000) = 1/10. So lambda' = 0.1*lambda.
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