Answer to Question #83011 in Quantum Mechanics for Niharika

Question

an electron is confined in one dimensional well of width 3*10 power of -10 meter.Find the kinetic energy of electron when it is the ground state

Answer 1

The Schrödinger equation for stationary states in the case of a one-dimensional problem is written in the form:

(∂^2 x)/(∂x^2 )+2m/ħ^2 (E-U)=0, boundary conditions have the form: Ψ(0)=Ψ(l)=0, the Schrödinger equation - (∂^2 x)/(∂x^2 )+2m/ħ^2 EΨ=0 or (∂^2 x)/(∂x^2 )+k^2 Ψ=0, where k^2=2mE/ħ^2

The general solution of a differential equation: Ψ(x)=A sin⁡〖kx+B cos⁡kx 〗, but Ψ(0)=0, then Ψ(x)=A sin⁡〖kx→Ψ(l)=A sin⁡〖kl=0〗 〗→kl=πn→k=πn/l

E_n=n^2 (π^2 ħ^2)/(2ml^2 ), where l = 3×〖10〗^(-10) m, n =1 because electron in the main state.

m=9.1×〖10〗^(-31) kg, ħ=1.5×〖10〗^(-34) J∙s

E_n=1^2 (π^2 (1.5×〖10〗^(-35) )^2)/(2∙9.1×〖10〗^(-31)∙(3×〖10〗^(-10) )^2 )≈6.7×〖10〗^(-19) J≈4.1 eV

The Schrödinger equation for stationary states in the case of a one-dimensional problem is written in the form:

(∂^2 x)/(∂x^2 )+2m/ħ^2 (E-U)=0, boundary conditions have the form: Ψ(0)=Ψ(l)=0, the Schrödinger equation - (∂^2 x)/(∂x^2 )+2m/ħ^2 EΨ=0 or (∂^2 x)/(∂x^2 )+k^2 Ψ=0, where k^2=2mE/ħ^2

The general solution of a differential equation: Ψ(x)=A sin⁡〖kx+B cos⁡kx 〗, but Ψ(0)=0, then Ψ(x)=A sin⁡〖kx→Ψ(l)=A sin⁡〖kl=0〗 〗→kl=πn→k=πn/l

E_n=n^2 (π^2 ħ^2)/(2ml^2 ), where l = 3×〖10〗^(-10) m, n =1 because electron in the main state.

m=9.1×〖10〗^(-31) kg, ħ=1.5×〖10〗^(-34) J∙s

E_n=1^2 (π^2 (1.5×〖10〗^(-35) )^2)/(2∙9.1×〖10〗^(-31)∙(3×〖10〗^(-10) )^2 )≈6.7×〖10〗^(-19) J≈4.1 eV