Question

o practice Problem-Solving Strategy 39.1: Particles and Waves. A molecule of hydrogen gas has a mass of 3.35×10−27kg3.35×10−27kg and a diameter of 1.48×10−10m1.48×10−10m. What is the kinetic energy at which this molecule's de Broglie wavelength will be equal to its diameter

Answer 1

0.0187 eV

Step-by-step explanation:

Given that:

diameter of the hydrogen gas (λ) = 1.48 ×10⁻¹⁰ m'

mass of the hydrogen gas = 3.35 ×10⁻²⁷ kg

We need to determine the momentum first before calculating the kinetic energy.

So momentum of the hydrogen gas molecule is written as;

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

`p = (96.626*10^(-34)J.s)/(1.48*10^(-10)m)`

`p=4.477*10^(-24) kg.m/s`

NOW, the kinetic energy of the hydrogen gas molecule is calculated as follows by using the formula:

`k_o = (p^2)/(2m)`

`k_o =\frac{(4.477*10^(-24)kg.m/s)^2}{2(3.35*10^-{27})kg}`

`k_o=2.9916*10^(-21)J((1eV)/(1.6*10^(-19)J))`

`K_o=0.0187eV`