From: Andrew Lorimer Date: Wed, 22 Aug 2018 23:52:02 +0000 (+1000) Subject: retarding voltage of electrons X-Git-Tag: yr11~59 X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/9740627fac3bbcc7b442821fc1e6d3eafca7273b retarding voltage of electrons --- diff --git a/physics/light-matter.md b/physics/light-matter.md index 940605b..19c50c3 100644 --- a/physics/light-matter.md +++ b/physics/light-matter.md @@ -10,7 +10,7 @@ $$\therefore E={hc \over \lambda}$$ where $E$ is energy of a quantum of light (J) $f$ is frequency of EM radiation -$h$ is Planck's constant ($6.63 \times 10^{-34}\operatorname{J s}$) +$h$ is Planck's constant ($6.63 \times 10^{-34}\operatorname{J s}=4.12 \times 10^{-15} \operatorname{eV s}$) ### Electron-volts @@ -68,12 +68,18 @@ $\phi$ is work function ("latent" energy) Gradient of a frequency-energy graph is equal to $h$ y-intercept is equal to $\phi$ +#### Stopping potential $V_0$ +$$V_0 = {E_{K \operatorname{max}} \over q_e} = {{hf - \phi} \over q_e}$$ + ## Wave-particle duality ### Double slit experiment Particle model allows potential for photons to interact as they pass through slits. However, an interference pattern still appears when a dim light source is used so that only one photon can pass at a time. ## De Broglie's theory + +$$\lambda = {h \over \rho} = {h \over mv}$$ + - theorised that matter may display both wave- and particle-like properties like light - predict wavelength of a particle with $\lambda = {h \over \rho}$ where $\rho = mv$ - impossible to confirm de Broglie's theory of matter with double-slit experiment, since wavelengths are much smaller than for light, requiring an equally small slit ($< r_{\operatorname{proton}}$) @@ -85,6 +91,8 @@ Particle model allows potential for photons to interact as they pass through sli - if $2\pi r \ne n{h \over mv}$, interference occurs when pattern is looped and standing wave cannot be established ### Photon momentum + +$$\rho = {hf \over c} = {h \over \lambda}$$ - if a massy particle (e.g. electron) has a wavelength, then anything with a wavelength must have momentum - therefore photons have (theoretical) momentum - to solve photon momentum, rearrange $\lambda = {h \over mv}$