A) 2.25 eV
B) 3.25 eV
C) 4.24 eV
D) 5.25 eV
Correct Answer: B
Solution :
Comparing the given equation of light with the equation \[E={{E}_{0}}\]\[(sin{{\omega }_{1}}t+sin{{\omega }_{2}}t)\]we get \[{{\omega }_{1}}=6\times {{10}^{15}}\]and \[{{\omega }_{2}}=8\times {{10}^{15}}\] \[{{v}_{1}}\frac{{{\omega }_{1}}}{2\pi }=\frac{6\times {{10}^{15}}}{2\pi }\]and \[{{v}_{2}}=\frac{{{\omega }_{2}}}{2\pi }=\frac{8\times {{10}^{15}}}{2\pi }Hz.\] \[\therefore \]Maximum frequency, \[{{v}_{2}}=\frac{9\times {{10}^{15}}}{2\pi }Hz.\] \[{{(KE)}_{\max }}=h{{v}_{2}}-{{\phi }_{0}}\] \[=\frac{66\times {{10}^{-34}}}{1.6\times {{10}^{-19}}}\times \frac{8\times {{10}^{15}}}{2\pi }-2=5.25-2=3.25\,eV\] Hence, the correction option is (b).
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