The smallest atoms can themselves exhibit quantum-mechanical behavior. Calculate the de Broglie wavelength (in picometers) of a hydrogen atom traveling at 490 m/s. Express your answer using three significant figures.

810 pm

Explanation:

Step 1: Given and required data

Velocity of the atom (v): 490 m/sMass of a hydrogen atom (m): 1.67 × 10⁻²⁷ kgPlanck's constant (h): 6.63 × 10⁻³⁴ J.s

Step 2: Calculate the de Broglie wavelength of the hydrogen atom

We will use de Broglie's equation.

λ = h / m × v

λ = 6.63 × 10⁻³⁴ J.s / 1.67 × 10⁻²⁷ kg × 490 m/s = 8.10 × 10⁻¹⁰ m

Step 3: Convert 8.10 × 10⁻¹⁰ m to picometers

We will use the conversion factor 1 m = 10¹² pm.

8.10 × 10⁻¹⁰ m × 10¹² pm/1 m = 810 pm

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.