A) \[5.2\,\times \,{{10}^{-6}}T\]
B) \[4.0\,\times \,{{10}^{-4}}T\]
C) \[4.0\,\times \,{{10}^{-6}}T\]
D) \[5.2\,\times \,{{10}^{-4}}T\]
Correct Answer: A
Solution :
Key Idea: Body performing circular motion is always acted upon by a centripetal force provided by magnetic field. If the kinetic energy of photoelectrons emitted from the metal surface is \[{{E}^{1}}\]and W the work function then from Einstein?s photoelectric equation, we have\[{{E}_{k}}=hv-W\]where v is frequency and h the Planck's constant \[\therefore \] \[\frac{1}{2}mv_{\max }^{2}=\frac{hc}{\lambda }-W\] \[v_{\max }^{2}=\frac{2}{m}\left[ \frac{hc}{\lambda }-W \right]\] Putting the numerical values from the question, we have \[\therefore \] \[v_{\max }^{2}=\frac{2}{9.1\times {{10}^{-31}}}\] \[\times \left[ \frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4\times {{10}^{-7}}}-2.5\times 1.6\times {{10}^{-19}} \right]\] \[=\frac{2}{9.1\times {{10}^{-31}}}\left[ 4.95\times {{10}^{-19}}-4\times {{10}^{-19}} \right]\] \[=0.208\times {{10}^{12}}\] \[\therefore \] \[{{v}_{\max }}=4.6\times {{10}^{15}}\,m/s\] A body performing circular motion is acted upon by a force directed towards its centre \[F=\frac{mv_{\max }^{2}}{r}\] Also the magnetic force \[F=e{{v}_{\max }}B\] On equating, we have \[\frac{mv_{\max }^{2}}{r}=e{{v}_{\max }}B\] \[\Rightarrow \]\[B=\frac{m{{v}_{\max }}}{er}=\frac{9.1\times {{10}^{-31}}\times 4.6\times {{10}^{5}}}{1.6\times {{10}^{-19}}\times 50\times {{10}^{-2}}}\] \[=5.2\times {{10}^{-6}}T\]
You need to login to perform this action.
You will be redirected in
3 sec
