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

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