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A particle (m = 3. 5 × 10-28 kg) starting from rest, experiences an acceleration of 2. 4 × 107 m/s2 for 5. 0 s. what is its de broglie wavelength λ at the end of this period?

Respuesta :

tanisharawat014

The de Broglie wavelength of the particle is 1.578*10^-3nm.

To find the answer, we have to know about the de Broglie wavelength.

How to find the de Broglie wavelength?

  • We have the expression for de Broglie wavelength as,

                          [tex]wavelength=\frac{h}{P} =\frac{h}{mv}[/tex]

where, h is the plank's constant, m is the mass and v is the velocity.

  • It is given that,

                               [tex]m=3.5*10^{-28}kg\\a=2.4*10^7m/s^2\\t=5s\\h=6.63*10^{-34}Js[/tex]

  • Substituting the values, we get,

                           [tex]wavelength=\frac{6.63*10^{-34}}{3.5*10^{-28}*2.4*10^7*5}=1.578*10^{-14}m\\\\ wavelength=1.578*10^{-3}nm.[/tex]

Thus, we can conclude that, the de Broglie wavelength of the particle is 1.578*10^-3nm.

Learn more about the the de Broglie wavelength here:

https://brainly.com/question/16595523

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