# Quantum Chemistry

Chapter06 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41

Exercise. This is the solution to exercise 6.1 in the book.

Solution.

$\psi \left(x,y,z\right)={A}_{x}{A}_{y}{A}_{z}sin\frac{{n}_{x}\pi x}{a}sin\frac{{n}_{y}\pi y}{b}sin\frac{{n}_{z}\pi z}{c}$

We must have that

${\int }_{0}^{a}{\int }_{0}^{b}{\int }_{0}^{c}{\psi }^{\ast }\left(x,y,z\right)\psi \left(x,y,z\right)dxdydz=1$

If we assign

${A}_{x}{A}_{y}{A}_{z}=C$

then

${C}^{2}{\int }_{0}^{a}{sin}^{2}\frac{{n}_{x}\pi x}{a}dx{\int }_{0}^{b}{sin}^{2}\frac{{n}_{y}\pi y}{b}dy{\int }_{0}^{c}{sin}^{2}\frac{{n}_{z}\pi z}{c}dz=1$

or

${C}^{2}\frac{a}{2}\frac{b}{2}\frac{c}{2}=1$

Hence

$C=\sqrt{\frac{8}{abc}}$

$\psi \left(x,y,z\right)=\sqrt{\frac{8}{abc}}sin\frac{{n}_{x}\pi x}{a}sin\frac{{n}_{y}\pi y}{b}sin\frac{{n}_{z}\pi z}{c}$