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

Wavelength,
`\lambda=1.28* 10^(-14)\ m`

Step-by-step explanation:

Given that,

Mass of the particle,
`m=4.3* 10^(-28)\ kg`

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

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

`\lambda=(h)/(mv)`

v = a × t

`\lambda=(h)/(mat)`

`\lambda=(6.67* 10^(-34))/(4.3* 10^(-28)* 2.4* 10^7* 5)`

`\lambda=1.28* 10^(-14)\ m`

Hence, this is the required solution.