A) \[6\times {{10}^{11}}\]
B) \[9\times {{10}^{11}}\]
C) \[11\times {{10}^{11}}\]
D) \[15\times {{10}^{11}}\]
Correct Answer: B
Solution :
Given, \[{{\lambda }_{1}}=4972{A}\]and \[{{\lambda }_{2}}=6216{A}\] and \[I=3.6\times {{10}^{-3}}W{{m}^{-2}}\] Intensity associated with each wavelength \[=\frac{3.6\times {{10}^{-3}}}{2}\]\[=1.8\times {{10}^{-3}}W{{m}^{-2}}\] work function \[\phi =hv\]\[=\frac{hc}{\lambda }\] \[=\frac{\left( 6.62\times {{10}^{-34}} \right)\left( 3\times {{10}^{8}} \right)}{\lambda }\]\[=\frac{12.4\times {{10}^{3}}}{\lambda }ev\] for different wavelengths \[\phi {{ }_{1}}=\frac{12.4\times {{10}^{3}}}{{{\lambda }_{1}}}=\frac{12.4\times {{10}^{3}}}{4972}=2.493eV\] \[=3.984\times {{10}^{-19}}J\] \[\phi {{ }_{2}}=\frac{12.4\times {{10}^{3}}}{{{\lambda }_{2}}}=\frac{12.4\times {{10}^{3}}}{6216}=1.994eV\] \[=3.184\times {{10}^{-19}}J\] Work function for metallic surface \[\phi =2.3eV\] (given) \[{{\phi }_{2}}<\phi \] Therefore, \[{{\phi }_{2}}\]will not contribute in this process. Now, no. of electrons per \[{{m}^{2}}-s=\] no. of photons per \[{{m}^{2}}-s\] no. of electrons per \[{{m}^{2}}-s=\frac{1.8\times {{10}^{-3}}}{3.984\times {{10}^{-19}}}\times {{10}^{-4}}\] \[\left( \because 1c{{m}^{2}}={{10}^{-4}}{{m}^{2}} \right)\]\[=0.45\times {{10}^{12}}\] So, the number of photo electrons liberated in 2 sec. \[=0.45\times {{10}^{12}}\times 2\] \[=9\times {{10}^{11}}\]
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