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A stream of electrons from a heated filament is accelerated through a potential difference of V esu between two charged plates. If e is the charge and m is the mass of an electron, what is h/lambda, where lambda is the de Broglie wavelength associated with the electron?

1. m*e*V
2. 2*m*e*V
3. sqrt(m*e*V)
4. sqrt(2*m*e*V)

Correct answer: sqrt(2*m*e*V)

Solution

An electron accelerated through potential V gains kinetic energy: (1/2)*m*v² = e*V. Hence v = sqrt(2eV/m). Momentum p = m*v = m*sqrt(2eV/m) = sqrt(2meV). Since lambda = h/p, we have h/lambda = p = sqrt(2meV).

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