Question

A particle (m = 3.8 × 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?

Answer 1

Answer: The De broglie's wavelength is 1.45×10^-14m

Step-by-step explanation:

Using De broglie's equation

λ= h/mv

Where h= Planck's constant = 6.6×10^-34Js

m= mass in kilogram = 3.8×10^-28kg

V= velocity = acceleration ×time =

2.4×10^7 ×5

V= 1.2×10^8m/s

λ= is the De broglie's wavelength

λ= 6.6×10^-34Js /(3.8×10^-28kg ×1.2×10^8)

λ= 1.45×10^-14m

Therefore ,De broglie's wavelength=

1.45×10^-14m