• question_answer An $\alpha -particle$and proton are accelerated from the state of rest through the same Potential difference, V. Find the ratio of de-Broglie wavelengths associated with them. Or If light of wavelength $\lambda =4000\overset{{}^\circ }{\mathop{A}}\,$ and intensity $100\text{ }W/{{m}^{2}}$ incident on a metal plate of threshold frequency $5.5\times {{10}^{14}}\text{ }Hz.$ What will be the maximum kinetic energy of photoelectron and number of photons incident per ${{m}^{2}}$ per second. $(h=6.6\times {{10}^{-34}}Js).$

De-Broglie wavelength, $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2m{{E}_{K}}}}'$ where p is momentum and ${{E}_{K}}$ is kinetic energy of a particle. For charged particle accelerated through a potential difference V, kinetic energy of particle                         ${{E}_{K}}=qV.$ $\therefore$ de-Broglie wavelength, $\lambda =\frac{h}{\sqrt{2mqV}}$ For $\alpha -particle,$ $m=4{{m}_{p}},$ q = 2e $\therefore$      ${{\lambda }_{\alpha }}=\frac{h}{\sqrt{2\times 4{{m}_{p}}\times 2eV}}=\frac{h}{4\sqrt{{{m}_{p}}eV}}$ For Proton, $m={{m}_{p}},$ q = e $\therefore {{\lambda }_{p}}=h/\sqrt{2{{m}_{p}}\,\,eV}$ $\therefore$ Ratio of de-Broglie wavelengths, $\frac{{{\lambda }_{\alpha }}}{{{\lambda }_{p}}}=\frac{1}{\sqrt{8}}$ $\therefore$      ${{\lambda }_{\alpha }}:{{\lambda }_{p}}=1:2\sqrt{2}$ Or Given,   $\lambda =4000\times {{10}^{-10}}m=4\times {{10}^{-7}}m$ ${{v}_{0}}=5.5\times {{10}^{14}}Hz$ Intensity, $I=100W/{{m}^{2}}.$ Suppose, n be the number of photons incident per ${{m}^{2}}$per second, then $I=nhc/\lambda$ or $n=\frac{I\lambda }{hc}=\frac{100\times (4\times {{10}^{-7}})}{(6.6\times {{10}^{-34}})\times (3\times {{10}^{8}})}=2.02\times {{10}^{20}}$ As, maximum kinetic energy of photoelectrons is ${{E}_{K}}=\frac{hc}{\lambda }-{{\phi }_{0}}=\frac{hc}{\lambda }-h{{v}_{0}}$ $\Rightarrow {{E}_{K}}=\frac{(6.6\times {{10}^{-34}})\times (3\times {{10}^{8}})}{4\times {{10}^{-7}}}$$-(6.6\times {{10}^{-34}})\times (5.5\times {{10}^{14}})$ $=1.32\times {{10}^{-19}}J$