Formula: Photoelectric Effect Energy Frequency Work function Frequency Velocity
$$h \, \class{violet}{f} ~=~ \frac{1}{2} \, m_{\text e} \, \class{blue}{v}^2 ~+~ \class{gray}{W}$$
$$W_{\text p} ~=~ \frac{1}{2} \, m_{\text e} \, \class{blue}{v}^2 ~+~ \class{gray}{W}$$
$$\class{violet}{f} ~=~ \frac{1}{h} \, \left(\frac{1}{2} \, m_{\text e} \, \class{blue}{v}^2 ~+~ \class{gray}{W}\right)$$
$$\class{gray}{W} ~=~ h\,\class{violet}{f} ~-~ \frac{1}{2} \, m_{\text e} \, \class{blue}{v}^2$$
$$\class{red}{f_0} ~=~ \class{violet}{f} ~-~ \frac{1}{2} \, \frac{m_{\text e}}{h} \, \class{blue}{v}^2$$
$$\class{blue}{v} ~=~ \sqrt{ \frac{ 2(h\,\class{violet}{f} - \class{gray}{W}) }{ m_{\text e} } }$$
Photon energy
$$ W_{\text p} $$ Unit $$ \mathrm{J} = \mathrm{Nm} = \frac{ \mathrm{kg} \, \mathrm{m^2} }{ \mathrm{s}^2 } $$
The energy of a photon is the sum of the kinetic energy (of an electron released with the help of the photon) and the work function of the illuminated material. From the photon energy, the frequency of the light can be determined:
$$ f~=~ \frac{ W_{\text p} }{ h } $$
Frequency
$$ f $$ Unit $$ \mathrm{Hz} = \frac{ 1 }{ \mathrm{s} } $$
Frequency of the used light source. If you multiply light frequency with Planck's constant \(h\), you obtain the energy \(W_{\text p}\) of a photon.
Work function
$$ \class{gray}{W} $$ Unit $$ \mathrm{J} $$
Work function is the energy that must be spent to knock an electron out of a solid (e.g. a metal plate). It is usually given in units of "eV" (electron volt) and can be calculated with the help of the cutoff frequency \( f_0 \):
$$ W = h\, f_0 $$
Cutoff frequency
$$ \class{red}{f_0} $$ Unit $$ \mathrm{Hz} = \frac{ 1 }{ \mathrm{s} } $$
Cutoff frequency is the minimum frequency of light necessary to knock out electrons. Multiply it by the Planck's constant \(h\) to obtain the work function: \(W = h \, f_0 \).
Velocity
$$ \class{blue}{\boldsymbol v} $$ Unit $$ \frac{\mathrm m}{\mathrm s} $$
Maximum speed of an electron knocked out by a photon. If the material is illuminated by photons with a larger photon energy \(W_{\text p}\) ( at constant work function \(W\)), then also the electron speed becomes larger.
Electron mass
$$ m_{\text e} $$ Unit $$ \mathrm{kg} $$
Rest mass of the electron. It is a physical constant with the value:
$$ m_{\text e} = 9.109 \cdot 10^{-31} \, \mathrm{kg} $$
Planck's Constant
$$ h $$ Unit $$ \mathrm{Js} = \frac{ \mathrm{kg} \, \mathrm{m}^2 }{ \mathrm{s} } $$
Planck's constant \( h \) is a physical constant from quantum mechanics and has the following exact value:
$$ h ~=~ 6.626 \, 070 \, 15 ~\cdot~ 10^{-34} \, \mathrm{Js} $$