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Question:

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:

25

60

50

75

Solution:

Let the initial kinetic energy of the particle be KE₁ and the final kinetic energy be KE₂.
Given that KE₂ = 16KE₁.
The de-Broglie wavelength (λ) is given by the equation:
λ = h/p, where h is Planck's constant and p is the momentum of the particle.
The momentum of a particle is related to its kinetic energy by:
p = √(2mKE), where m is the mass of the particle.
Therefore, the de-Broglie wavelength can be written as:
λ = h/√(2mKE)
Let λ₁ be the initial de-Broglie wavelength and λ₂ be the final de-Broglie wavelength.
Then:
λ₁ = h/√(2mKE₁)
λ₂ = h/√(2mKE₂) = h/√(2m(16KE₁)) = h/(4√(2mKE₁))
Now, we can find the ratio of the final wavelength to the initial wavelength:
λ₂/λ₁ = [h/(4√(2mKE₁))] / [h/√(2mKE₁)] = 1/4
This means that λ₂ = λ₁/4.
The percentage change in the de-Broglie wavelength is given by:
Percentage change = [(λ₂ - λ₁)/λ₁] × 100
Percentage change = [(λ₁/4 - λ₁)/λ₁] × 100 = (-3λ₁/4λ₁) × 100 = -75%
The negative sign indicates a decrease in the wavelength. Therefore, the percentage change in the de-Broglie wavelength is 75% decrease. The question asks for the percentage change, which is 75.