Question

A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de Broglie wavelength λ at the end of this period?

Answer 1

Answer:

Wavelength, $\lambda=1.28\times 10^{-14}\ m$

Explanation:

Given that,

Mass of the particle, $m=4.3\times 10^{-28}\ kg$

Acceleration of the particle, $a=2.4\times 10^7\ m/s^2$

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

$\lambda=\dfrac{h}{mv}$

v = a × t

$\lambda=\dfrac{h}{mat}$

$\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}$

$\lambda=1.28\times 10^{-14}\ m$

Hence, this is the required solution.