What is the percentage change in the de-broglie wavelength of the particle if its kinetic energy is increased by 15 times?

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what is the percentage change in the de-broglie wavelength of the particle if its kinetic energy is increased by 15 times?

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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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