Quantum Chemistry/Example 3

Question 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). $$

Part 1
nnodes=n-1 nnodes=2-1=1 We're looking for one x-intercept. $$ 0 \div \sqrt{\frac{2}{5}} = \sqrt{\frac{2}{5}}\sin\left( \frac{2\pi x}{5} \right)\div \sqrt{\frac{2}{5}}$$ $$ 0= \sqrt{\frac{2}{5}}\sin\left( \frac{2\pi x}{5} \right) $$ 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, $$\left( \frac{5}{2} \right)$$, 5 0 and 5 are the edges of the box and $$ \left( \frac{5}{2} \right) $$ is the only node.

Part 2
nnodes=n-1 nnodes=3-1=2 We're looking for two x-intercepts. $$ 0 \div \sqrt{\frac{3}{5}} = \sqrt{\frac{3}{5}}\sin\left( \frac{3\pi x}{5} \right)\div \sqrt{\frac{3}{5}}$$ $$ 0= \sqrt{\frac{3}{5}}\sin\left( \frac{3\pi x}{5} \right) $$ $$ k= \left( \frac{3\pi}{5} \right) $$ $$ Period= \left( \frac{3\pi}{\frac{3\pi}{5}} \right)= \frac{10}{3}$$ x-intercepts are $$\left( \frac{5n}{3} \right)$$ x-intercepts = 0, $$\left( \frac{5}{3} \right),\left( \frac{10}{3} \right),5 $$ 0 and 5 are the edges of the box, $$ \left( \frac{5}{3} \right) $$ and $$ \left( \frac{10}{3} \right) $$ are the nodes.