PARTICLES IN ONE DIMENSIONAL BOX

      Consider an electron move in one dimensional box with width L.

Let potential v=0 inside the box, v=& outside the box
Take Schrodinger time independent wave equation
d²ψ/dx² +2m(E-V)ψ/ħ²=0 . . . (1)
 Particles is in one dimensional box so, v=0
d²ψ/dx² +2mψE/ħ²=0 . . . . (2)
Let Ψ(x)=A sin kx+B cos kx . . . . (3)
be the general solution of eq(2)
boundary condition are x=0 at ψ(x)=0
apply boundary condition in eq(3)
0=A sink(0)+ B cos k(0)
0=0+B      .·.B=0
second boundary condition x=L at ψ(x)=0,B=0
apply the above condition to eq(3)
0=A sin kL+ B cos kL
 =A sin kL + 0 cos kL
 =A sin kL=0   A≠ 0
Asin kL =0
kL=±nπ, where n= 1 2 3 4 ..
k=nπ/L . . . . (4)
put eq(4)in eq (3) and B=0
ψ(x)=A sin nπ/L*x . . . (5)
now differentiate eq(5) with respect to x
(dψ/dx)x=A cos nπ/L*x *nπ/L
again differentiate with respect to x we get,
d²ψ/dx²= -nπ/L A sin nπ/L*x*nπ/L
d²ψ/dx² =-n²π²/L² A sin nπ/L*x
d²ψ/dx²=( -n²π²/L²)ψ
d²ψ/dx²+n²π²ψ/L²=0 . . . (6)
compare eq(6) with eq(2) we get
2mE/ħ²=n²π²/L²
E=n²π²ħ²/L²2m
E=n²π²h²/L²2m4π²
E=n²h²/L²8m . . . . (7)
where n is integral
           h is planks constant
           m is mass
           L is width

when n=1; E₁= h²/8mL²
          n=2; E₂=4h²/8mL²
          n=3; E₂=9h²/8mL²
Now find the constant value of A of eq(5) by normalization method
∫|ψ(x)|² dx=1                                                (we suppose the probability of finding electron is 1)
∫A²sin²(nπx/L)dx=1                                                                                     
A²∫1-cos(2nπx/L)/2 dx=1
A²/2∫[1-cos(2nπx/L)] dx=1
A²/2 x-sin(2nπx/L)/(2nπ/L)=1
A²/2[x-(L/2nπ) sin2nπx/L]=1   put x=L
A²/2 [L-0=1]
A²L/2 =1
A²=2/L  .·.A=√(2/L)
ψ(x)=√(2/L)sin nπx/L.