A) \[4.4\times {{10}^{5}}m/s\]
B) \[8.8\times {{10}^{7}}m/s\]
C) \[8.8\times {{10}^{5}}m/s\]
D) \[4.4\times {{10}^{7}}m/s\]
Correct Answer: B
Solution :
Work function \[W=4eV=4\times 1.6\times {{10}^{-19}}J\] \[=6.4\times {{10}^{-19}}J\] Wavelength of incident radiation \[\lambda =0.2\mu m=0.2\times {{10}^{-6}}m\] Maximum KE of liberated electron \[{{(KE)}_{\max }}=\frac{hc}{\lambda }-W\] \[\frac{1}{2}mv_{\max }^{2}=\frac{hc}{\lambda }-W\] \[\frac{1}{2}\times 9.1\times {{10}^{-31}}v_{\max }^{2}\] \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{0.2\times {{10}^{-6}}}-6.4\times {{10}^{-19}}\] \[=9.9\times {{10}^{-9}}-6.4\times {{10}^{-19}}=3.5\times {{10}^{-19}}\] \[\therefore \] \[{{v}_{\max }}=\sqrt{\frac{3.5\times {{10}^{-9}}\times 2}{9.1\times {{10}^{-31}}}}=\sqrt{\frac{7}{9.1}\times {{10}^{12}}}\]
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