# Quantum Chemistry/Example 3

Write a question about calculating the number of nodes in a particle in a 1D box wavefunction.

## 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).

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