Question

The strength of the transverse magnetic field required to bend all photoelectrons with in a circle of radius 50 cm when light of wavelength 4000 $ {{A}^{o}} $ is incident on barium emitter is (work function of barium is 2.5 eV):

• 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 $

Answer 1

Answer: (A)

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 $