Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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?

Sagot :

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

#SPJ4

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.