A) \[\text{5 }\times \text{ 1}{{0}^{\text{3}}}\text{ m}/\text{s}\]
B) \[\text{5 }\times \text{ 1}{{0}^{\text{4}}}\text{ m}/\text{s}\]
C) \[\text{5}.0\text{ }\times \text{ 1}{{0}^{\text{5}}}\text{ m}/\text{s}\]
D) \[\text{5 }\times \text{ 1}{{0}^{\text{6}}}\text{ m}/\text{s}\]
Correct Answer: C
Solution :
\[{{E}_{k}}=\frac{1}{2}m{{v}^{2}}\] \[\therefore \] \[{{v}_{\max }}=\sqrt{\frac{2{{E}_{K}}}{m}}\] Now, \[{{E}_{K}}=hv-{{W}_{0}}=\frac{hc}{\lambda }-{{W}_{o}}\] \[=\frac{6\cdot 63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{5000\times {{10}^{-10}}}-(2.9\times {{10}^{-19}}J)\] \[=(4.0-2.9)\times {{10}^{-10}}\mathbf{=1}\mathbf{.1\times 1}{{\mathbf{0}}^{\mathbf{-19}}}\mathbf{J}\] \[\therefore \]\[{{v}_{\max }}=\sqrt{\frac{2\times 1\cdot 1\times {{10}^{-19}}}{9\cdot 0\times {{10}^{-31}}}}=\mathbf{5}\mathbf{.0\times 1}{{\mathbf{0}}^{\mathbf{5}}}\mathbf{m/sec}\]
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