what is the percentage change in the de-broglie wavelength of the particle if its kinetic energy is increased by 15 times?
Answer: The de Broglie wavelength of a particle is given by the equation:
λ = h/p
where λ is the de Broglie wavelength, h is Planck’s constant, and p is the momentum of the particle.
The momentum of a particle is related to its kinetic energy E and mass m by the equation:
p = √(2mE)
So, if the kinetic energy of the particle is increased by a factor of 15, its momentum will also increase by a factor of √15. Therefore, the new momentum (p’) of the particle can be calculated as:
p’ = √(2m(15E))
p’ = √(30mE)
Substituting this expression for momentum into the de Broglie wavelength equation, we get:
λ’ = h/p’
λ’ = h/√(30mE)
So, the new de Broglie wavelength (λ’) is:
λ’ = λ/√15
Therefore, the percentage change in the de Broglie wavelength of the particle is:
[(λ’ - λ)/λ] x 100%
Substituting the expression for λ’ and simplifying, we get:
[(λ/√15 - λ)/λ] x 100%
= [(1/√15 - 1) x 100%]
≈ -18.26%
So, the percentage change in the de Broglie wavelength of the particle is approximately -18.26% if its kinetic energy is increased by 15 times.
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