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If the kinetic energy of a free electron is doubled, the new de Broglie wavelength will be ______ times the initial wavelength.

Asked in GUJCET 2014 · Same or changed energy

Answer: (2) 1/(√2)

Step-by-step solution

Given: K₂ = 2K₁ for the same electron.

Idea: λ = h/p = h/(√2mK), so λ ∝ 1/(√K).

(λ₂)/(λ₁) = √(K₁)/(K₂) = √1/2 = 1/(√2)

So the new wavelength is 1/(√2) times (about 0.71 times) the initial wavelength.

Why the other options are wrong

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