JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Important Formulae for Photoelectric Effect

(1) \[h\nu =h{{\nu }_{0}}+{{K}_{\max }}\] and \[{{K}_{\max }}=e{{V}_{0}}\]

(2) \[{{K}_{\max }}=e{{V}_{0}}=h(\nu -{{\nu }_{0}})\]\[\Rightarrow \]\[\frac{1}{2}mv_{\max }^{2}=h(\nu -{{\nu }_{0}})\]

(3) \[{{v}_{\max }}=\sqrt{\frac{2h(\nu -{{\nu }_{0}})}{m}}\]

(4) \[{{K}_{\max }}=\frac{1}{2}mv_{\max }^{2}=e{{V}_{0}}=hc\,\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=hc\,\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)\]

(5) \[{{v}_{\max }}=\sqrt{\frac{2hc}{m}\frac{\left( {{\lambda }_{0}}-\lambda  \right)}{\lambda {{\lambda }_{0}}}}\]

(6) \[{{V}_{0}}=\frac{h}{e}(\nu -{{\nu }_{0}})=\frac{hc}{e}\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=12375\,\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\]