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If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:

1. 25
2. 75
3. 6
4. 50

Correct answer: 75

Solution

The de-Broglie wavelength is inversely proportional to the square root of the kinetic energy (λ ∝ 1/√KE). If the kinetic energy increases by 16 times, the wavelength becomes 1/√16 = 1/4 of its original value. The percentage change is [(original - new)/original] × 100 = [(1 - 1/4)/1] × 100 = 75%.

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