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. \[(h=6.6\times {{10}^{-34}}Js).\]

Answer:

            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 \[{{E}_{K}}=\frac{hc}{\lambda }-{{\phi }_{0}}=\frac{hc}{\lambda }-h{{v}_{0}}\] \[\Rightarrow {{E}_{K}}=\frac{(6.6\times {{10}^{-34}})\times (3\times {{10}^{8}})}{4\times {{10}^{-7}}}\]\[-(6.6\times {{10}^{-34}})\times (5.5\times {{10}^{14}})\] \[=1.32\times {{10}^{-19}}J\]