1. An electron is trapped in a region between two perfectly rigid walls (which can be regarded as infinitely high energy barriers). In the region between the walls the potential energy of the electron is zero. The (normalized) wave function of the electron in the region between the walls is, (x) = A sin(bx) where A = 0.50 nm-1/2 and b = 1.18 nm-1/2 Sketch this problem showing the potential well, the given wave function, and its associated probability density. By substituting into the Schrödinger equation, find the energy of the electron in this state of motion. Calculate the probability for the electron to be found between x = 0.98 nm and x = 1.00 nm. 201 h* d*v 2m da² + U(x)(x) = E(2) h²n² En 8mL2 (n = 1,2,3,...) (x)=(mwo/h)/4-(√km/2h)x2 En=(n+hwo (n= 0, 1, 2, ...) aon² Tn= Ꮓ E₁ = -(13.6eV) Z2 n2 m= 4Teoh² me² = me4 1 -13.6 eV n² aon² (n = 1,2,3,...) AE memp me+mp dV = r² sin drdodo (10.2eV)(Z-1)² En 32m²² n² n² (n = 1,2,3,...) α= n² e2 Απελε 1 A = Alimit 12- n² (n =no +1, no +2, no + 3,...) AE = mc2a4- ηδ λ = 64x³ ³c n²n me4 1 = n²-n nin Ron-n P(r) = r²| Rn,1(r)|2 N(0) == =(*)(*) 2 1 sin 0/2 n(x)=√ 2 ппх sin L

1. An electron is trapped in a region between two perfectly rigid walls (which can be regarded as infinitely high energy barriers). In the region between the walls the potential energy of the electron is zero. The (normalized) wave function of the electron in the region between the walls is, (x) = A sin(bx) where A = 0.50 nm-1/2 and b = 1.18 nm-1/2 Sketch this problem showing the potential well, the given wave function, and its associated probability density. By substituting into the Schrödinger equation, find the energy of the electron in this state of motion. Calculate the probability for the electron to be found between x = 0.98 nm and x = 1.00 nm. 201 h* dv 2m da² + U(x)(x) = E(2) h²n² En 8mL2 (n = 1,2,3,...) (x)=(mwo/h)/4-(√km/2h)x2 En=(n+hwo (n= 0, 1, 2, ...) aon² Tn= Ꮓ E₁ = -(13.6eV) Z2 n2 m= 4Teoh² me² = me4 1 -13.6 eV n² aon² (n = 1,2,3,...) AE memp me+mp dV = r² sin drdodo (10.2eV)(Z-1)² En 32m²² n² n² (n = 1,2,3,...) α= n² e2 Απελε 1 A = Alimit 12- n² (n =no +1, no +2, no + 3,...) AE = mc2a4- ηδ λ = 64x³ ³c n²n me4 1 = n²-n nin Ron-n P(r) = r²| Rn,1(r)|2 N(0) == =()(*) 2 1 sin 0/2 n(x)=√ 2 ппх sin L