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What is the de Broglie wavelength associated with an electron accelerated through a potential difference of 64 V?

Asked in GUJCET 2023 · Accelerated through a potential difference

Answer: (3) 1.53 A

Step-by-step solution

Given: accelerating potential V = 64 V.

Idea: the electron gains kinetic energy K = eV, so λ = h/(√2meV), which works out to λ = (12.27)/(√V) A.

λ = (12.27)/(√64) = (12.27)/8 = 1.53 A

Check directly: √2×9.1×10⁻³¹×1.6×10⁻¹⁹×64 = 4.32×10⁻²⁴ kg m s⁻¹, and (6.63×10⁻³⁴)/(4.32×10⁻²⁴) = 1.53×10⁻¹⁰ m.

So λ = 1.53 A.

Why the other options are wrong

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