Quantum Chemistry/Example 3

Calculate the number of nodes for a particle in a 1D box when n=2 and n=3 when the length of the box is L=5, and give the x-intercept of the node(s).

Wavefunction for a particle in a 1D-Box $\psi (x)=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)$

Answer

Part 1

n_nodes=n-1

n_nodes=2-1=1

We're looking for one x-intercept.

$0÷\sqrt{\frac{2}{5}}=\sqrt{\frac{2}{5}}\sin \left(\frac{2\pi x}{5}\right)÷\sqrt{\frac{2}{5}}$

$0=\sqrt{\frac{2}{5}}\sin \left(\frac{2\pi x}{5}\right)$

Use the form $\sin (kx)$ where $k=\left(\frac{2\pi }{5}\right)$

$Period=\left(\frac{2\pi }{k}\right)$

$Period=\left(\frac{2\pi }{\frac{2\pi }{5}}\right)=5$

$\left(\frac{Period}{2}\right)$ is the x-intercept

x-intercepts are 0, $\frac{5}{2}$, 5

0 and 5 are the edges of the box and $\frac{5}{2}$ is the only node.

Part 2

n_nodes=n-1

n_nodes=3-1=2

We're looking for two x-intercepts.

$0÷\sqrt{\frac{3}{5}}=\sqrt{\frac{3}{5}}\sin \left(\frac{3\pi x}{5}\right)÷\sqrt{\frac{3}{5}}$

$0=\sqrt{\frac{3}{5}}\sin \left(\frac{3\pi x}{5}\right)$

$k=\left(\frac{3\pi }{5}\right)$ $Period=\left(\frac{3\pi }{\frac{3\pi }{5}}\right)=\frac{10}{3}$

x-intercepts are $\left(\frac{5n}{3}\right)$

x-intercepts = 0, $\frac{5}{3},\frac{10}{3},5$

0 and 5 are the edges of the box, $\frac{5}{3}$ and $\frac{10}{3}$ are the nodes.