• # question_answer Energy required to remove an electron from aluminium surface is 4.2 eV. If light of wavelength 2000 A falls on the surface, the velocity of the fastest electron ejected from the surface will be A) $8.4\times {{10}^{5}}$m/second B) $7.4\times {{10}^{5}}$m/second C) $6.4\times {{10}^{5}}$m/second D)  $8.4\times {{10}^{6}}$m/second

Solution :

: For photoelectric effect, $\frac{hc}{\lambda }=\phi +\frac{1}{2}m{{v}^{2}}$or$\frac{1}{2}m{{v}^{2}}=\frac{hc}{\lambda }-\phi$ $=\frac{(6.6\times {{10}^{-34}})\times (3\times {{10}^{8}})}{2000\times {{10}^{-10}}}-\frac{4.2\times 1.6\times {{10}^{-19}}}{1}$ Or $\frac{1}{2}\times (9\times {{10}^{-31}})\times {{v}^{2}}$ $=9.9\times {{10}^{-19}}-6.72\times {{10}^{-19}}$ $=3.18\times {{10}^{-19}}$ Or ${{v}^{2}}=\frac{2\times 3.18\times {{10}^{-19}}}{9\times {{10}^{-31}}}=70.66\times {{10}^{10}}$ Or $v=8.4\times {{10}^{5}}m/s$

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