Question

Determinati expresia razei orbitei electronului in atomul de hidrogen

Answer 1

Fcf=Fcp

m₀v²/r=e²/4πε₀r² |*m₀=>m₀²v²/r=m₀e²/4πε₀r² |*r³=>m₀²v²r²=m₀e²r/4πε₀=n²h²/4π²<=>m₀e²r/ε₀=n²h²/π=>r=n²h²ε₀/πm₀e²

n∈{1,2,3,..,7}->numarul cuantic principal

m₀=9,1*10⁻³¹ kg->masa electronului

ε₀=8,854*10⁻¹² F/m ->permitivitatea absolută a vidului

h=6,62*10⁻³⁴ Js->constanta lui Planck

e=1,6*10⁻¹⁹ C->sarcina electrica

r=n²h²ε₀/πm₀e²