## Photoelectric effect
+
+
### Planck's equation
$$E=hf,\quad f={c \over \lambda}$$
- therefore, stable orbits are those where circumference = whole number of e- wavelengths
- if $2\pi r \ne n{h \over mv}$, interference occurs when pattern is looped and standing wave cannot be established
+
+
### Photon momentum
$$\rho = {hf \over c} = {h \over \lambda}$$
- Black lines in spectrum show light not reflected
### Emission
+
+
+
- Coloured lines show light being ejected from e- shells
- Energy change between ground / excited state: $\Delta E = hf = {hc \over \lambda}$
- Bohr's model describes discrete energy levels
- EMR is absorbed/emitted when $E_{\operatorname{K-in}}=\Delta E_{\operatorname{shells}}$ (i.e. $\lambda = {hc \over \Delta E_{\operatorname{shells}}}$)
## Light sources
+
+
+
- **incandescent:** <10% efficient, broad spectrum
- **LED:** semiconducting doped-Si diodes
- - most electrons in *valence band* (energy level)
| ------------------------------- | ---------------------------- |
| $k$ (constant) | $kx + c$ |
| $x^n$ | ${1 \over {n+1}}x^{n+1} + c$ |
-| $a \cdot {1 \over x}$ | $a \cdot \log_e x + c$ |
+| $a x^{-n}$ | $a \cdot \log_e x + c$ |
| $e^{kx}$ | ${1 \over k} e^{kx} + c$ |
| $e^k$ | $e^kx + c$ |
| $\sin kx$ | $-{1 \over k} \cos (kx) + c$ |
| ${f^\prime (x)} \over {f(x)}$ | $\log_e f(x) + c$ |
| $g^\prime(x)\cdot f^\prime(g(x)$ | $f(g(x))$ (chain rule)|
| $f(x) \cdot g(x)$ | $\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx$ |
+| ${1 \over {ax+b}}$ | ${1 \over a} \log_e (ax+b) + c$ |
+| $(ax+b)^n$ | ${1 \over {a(n+1)}}(ax+b)^{n-1} + c$ |
+
+## Applications of antidifferentiation
+
+- $x$-intercepts of $y=f(x)$ identify $x$-coordinates of stationary points on $y=F(x)$
+- the nature of any stationary point of $y=F(x)$ is determined by the way the sign of the graph of $y=f(x)$ changes about its $x$-intercepts
+- if $f(x)$ is a polynomial of degree $n$, then $F(x)$ has degree $n+1$
+
+To find stationary points of a function, substitute $x$ value of given point into derivative. Solve for ${dy \over dx}=0$. Integrate to find original function.
