**¥**

The particle cannot be at locations, x > L or less than 0, else it would have infinite energy.

**Therefore, outside the interval (0,L) one has the quantum wave form: y*y = 0**

**The Schrodinger equation to solve is: dy**

where K = Ö [2mE] / ħ

And this second order differential equation has the solution:

y = C sin Kx + D cos Kx

^{2}/dx^{2}+ K^{2}y = 0where K = Ö [2mE] / ħ

And this second order differential equation has the solution:

y = C sin Kx + D cos Kx

**(C, D constants)**

We set boundary conditions thus:

We set boundary conditions thus:

**At x = 0+, y (0+) = D (since sin(0) = 0)**

**At x = 0, y (0) = 0**

**The wave function must be continuous (since we want y*y to represent something physically observable).**

So: y (0+) = y (0-) therefore D = 0

y = C sin Kx

Now, at x = L+, y (L+) = 0

So: y (0+) = y (0-) therefore D = 0

y = C sin Kx

Now, at x = L+, y (L+) = 0

**At x = L-, y**

**(L-) = C sin K(L)**

*For continuity*: y (L+) = y**(L-)**

**Therefore: C sin K(L-) = 0**

__but C can't = 0__, or no particle!

**Therefore: sin K(L-) = 0 Þ sin (Ö**

**[2mE] / ħ) L = 0,**

**and**

**((Ö**

**[2mE] / ħ) L = n π**

**Therefore:**

**L = nπ /K and n = + 1, + 2, ..etc.**

**2mE = n**

^{2}π^{2}ħ^{2 }Then**E = { n**

From which we obtain the energy eigenvalues for the levels n = 1 and n=2.

^{2}π^{2}ħ^{2}} / 2mL^{2}From which we obtain the energy eigenvalues for the levels n = 1 and n=2.

**For n = 1: E**

_{n=1}= π^{2}ħ^{2}/ 2mL^{2},

**For energy level n = 2: E**

_{n=2}= 4 π^{2}ħ^{2}/ 2mL^{2}
Now m = 9.1 x 10

^{-31}kg (mass of electron) and L = 0.2nm = 2 x 10^{-10}m
Also:

**ħ = h/ 2 π**= 6.62 x 10^{-34}J-s/ 2 π = 1.054 x 10^{-34}J-s
Then the actual energy at the first level (n= 1) is:

**π**= (3.1416)

^{2}ħ^{2}/ 2mL^{2 }^{2}(1.054 x 10

^{-34}J-s)

^{2}

**/**2(9.1 x 10

^{-31}kg) (2 x 10

^{-10}m)

^{ 16}J

Then the actual energy at the 2nd level (n= 2) is:

**4 E**

_{n=1}=**4(**1.43 x 10

^{ 16}J) = 5.72 x 10

^{ 16}J

**(2)**What is the probability the particle will be found between x = 0 and x = L/4.

A particle confined in a box of length L has a wave function that can be expressed in simplest form as:

y = Ö2/ ÖL [sin (2npx/L)]

y = Ö2/ ÖL [sin (2npx/L)]

Note that if: y = C sin Kx

**then normalization requires we do the following:**

**ò**

^{L}_{o}_{ }‖y‖

^{2 }dx

**ò**

^{L}_{o}_{ }‖C‖

^{2 }sin

^{2}Kx

**dx**

^{2}L/ 2 = 1

So that: ‖C‖

^{2}= 2/L and C = Ö2/ ÖLConsider the simplest energy level : n = 1

Then: P

_{ab}_{ }=**ò**^{L/4 }_{0}_{ }(Ö2/ ÖL )^{2}^{ }sin^{2}(2px/L) dxLet q = 2px/L and use the trig identity: sin

^{2}q= ½ (1 – cos 2q)

P

_{ab }= 2/ L**ò**^{L/4 }_{0}_{ }½ [1 - cos (4px/L)] dx
P

_{ab }= 2/ L [½**ò**^{L/4 }_{0}_{ }dx -**ò**^{ L/4 }_{0}_{ }cos (4px/L)] dx

And: P

_{ab }= x/ L - 1/2p sin (4px/L)]

^{ L/4 }

_{0 }

P

_{ab }= ¼ - 0 = 0.25
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