- +44 141 628 6080
- info@tutorsglobe.com

18,76,764

Questions

Asked

21,311

Experts

9,67,568

Questions

Answered

Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

Submit Assignment
## Particle in Three-Dimensional Box, Chemistry tutorial

IntroductionSuppose we consider the similar particle as in chapter but this time it is constrained to shift in a rectangular box of dimensions a, b and c in length. Within the box (i.e. between x= 0 and a; y = 0 and b and z = 0 and c), the potential energy is zero. At the walls and everywhere outside the box, the potential is.

Definition of Particle in a Three-Dimensional BoxThe Schrodinger wave equation for 3 dimensional (3D) box is

δ

^{2}φ/δx^{2}+ δ^{2}φ/δy^{2}+ δ^{2}φ/δz^{2}+ 8 π^{2}m/h^{2}(E-V) φ =0where and V are (x,y,z).

since V = 0 inside the box, then the equation becomes

δ

^{2}φ/δx^{2}+ δ^{2}φ/δy^{2}+ δ^{2}φ/δz^{2}+ 8 π^{2}m/h^{2}Eφ=0Equation may be solved by writing the wave function as the product of 3 functions, each depending on one coordinate.

φ(x, y, z) = X(x)Y(y)Z(z)

differentiating equation gives

δφ/δx = Y (y) Z (z) δx/δx

δ

^{2}φ/δx^{2}= Y (y) Z(z) δ^{2}x/δx^{2}and by a similar reasoning

δ

^{2}φ/δy^{2}= X (x)Z(z) δ^{2}y/δy^{2}δ

^{2}φ/δz^{2}= X(x)Y(y) δ^{2}z/δz^{2}Y(y)Z(z)δ

^{2}x/δx^{2}+ X(x)Z(z) ) δ^{2}y/δy^{2}+ X(x)Y(y) δ^{2}z/δz^{2}+ 8 π^{2}m/h^{2}EX(x)Y(y)Z(z) = 0 dividing all through by Y(y)X(x)Z(z) one obtains-h^{2}/8 π^{2 }m* 1/X(x) δ^{2}x/δx^{2 }+1/Y(y) * δ^{2}y/δy^{2}^{ }+ 1/Z(z) δ^{2}z/δz^{2 }= EWe can write the energy as the sum of three contributions associated with the coordinates.

E = E

_{x}+ E_{y }+ E_{z}using equation, we can divide the expression attained into 3 equations.

-h

^{2}/8 π^{2 }m *1/x δ^{2}x/δx^{2 }= E_{x}-h

^{2}/8 π^{2 }m *1/y δ^{2}y/δy^{2}= E_{y }-h

^{2}/8 π^{2 }m *1/z δ^{2}z/δz^{2}= E_{z}Each of the last three equations is similar to the expression for the particle in a one-dimensional box discussed in chapter. Hence their solutions are respectively:

X = [√(

^{2}/a)]sin nπx/aEn,x = n

^{2}_{x}h^{2}/8ma^{2}Y = [√(2/b)]sin nπy/b

E

_{n,y }= n^{2}_{y }h^{2}/ 8mb^{2}X = [√(

^{2}/c)] sin nπx/cE

_{n,z }= n^{2}_{z }h^{2}/ 8mc^{2}Where a, b, c are length in x, y, z directions, correspondingly. Also, nx, ny, nz are the quantum numbers correspondingly.

Since φ (x, y, z) = X(x)Y(y)Z(z) and E = E

_{x}+ E_{y}+ E_{z}, thenψ (x,y,z) = √a/v sin(nπx/a) sin(nπy/b) sin(nπz/c)

where V is the volume of the box.

E

_{x,y,z }= h^{2}/δm * n^{2}_{x}/a^{2}+ n^{2}_{y}/b^{2}[+ n^{2}_{z}/c^{2}]Whenever the three dimensional box has geometrical symmetry, more interesting results are often obtained. For example, in a cubic box, a = b =c, thus equation becomes

E = h

^{2}/ φm(n^{2}_{x}+ n^{2}_{y }+ n^{2}_{z})Suppose n

_{x}=3, n_{y}= n_{z}=2, thenΨ(x,y,z) = [/(8/v)] sin3πx/a* sin2πy/a*sin2πz/a]

E = h

^{2}/8ma^{2}(3^{2}+ 2^{2}+ 2^{2}) = 17h^{2}/8ma^{2}Assuming we have another set of values n

_{x}=2, n_{y}=3, n_{z}=2, thenΨ(x,y,z) = [/(8/v )]sin 2πx/a sin3πy/a sin2πz/a

E = h

^{2}/8ma^{2}(2^{2}+ 3^{2}+ 2^{2}) = 17h^{2}/8ma^{2}Suppose n

_{x}=2, n_{y }=2, n_{z}=3, thenΨ(x,y,z) = [/( 8/v)]sin2πx/a sin2πy/a sin3πz/a

E = h

^{2}/8ma^{2}(2^{2}+ 2^{2}+ 3^{2}) = 17h^{2}/8ma^{2}Even though such states are different, the values of the energies are the same. The three states are said to be degenerate since they have equal energy.

For a situation where nx = ny = nz = 1, it corresponds to only one state of the system. The same is true of nx = ny = nz = 2, but situations such as nx ≠ ny = nz three degenerate states are obtained as shown in the Figure below.

Fig: Quantized Energy Levels of a Particle in a Cubic Box

Suppose we wish to calculate the transition energy between the levels E2,2.2 and E3,2,1 then the energy difference is calculated as

ΔE = hν = 14h

^{2}/8ma^{2}- 12h^{2}/8ma^{2}= 2h^{2}/8ma^{2}= h^{2}/4ma^{2}Given suitable data, it should be possible for us to calculate the frequency (ν) of the transition between the two states.

Zero point energyAccording the old quantum theory, the energy level of a harmonic oscillator is E = hv.

The lowest energy level via n = 0 would have zero energy. Depend on the wave treatment of the system, the energy level corresponds to the state with quantum numbers n

_{x }= n_{y}= n_{z}=1. The dissimilarity between such 2 values is termed the zero point energy.Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with an expert at http://www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online chemistry tutoring. Chat with us or submit request at info@tutorsglobe.com