Answer:
\[8\times {{10}^{6}}m/s\] \[K{{E}_{\max }}=h\text{v}-{{\phi }_{0}}\] Where, \[{{\phi }_{0}}\] is the work function. Given, \[{{\phi }_{0}}=h{{\text{v}}_{\text{0}}}\] \[\frac{1}{2}m\,{{(4\times {{10}^{6}})}^{2}}=2h{{\text{v}}_{\text{0}}}-h{{\text{v}}_{\text{0}}}\] i.e. \[\frac{1}{2}m\,{{(4\times {{10}^{6}})}^{2}}=h{{\text{v}}_{\text{0}}}\] ? (i) \[\frac{1}{2}m\,{{({{V}_{max}})}^{2}}=5h{{\text{v}}_{\text{0}}}-h{{\text{v}}_{\text{0}}}=4h{{\text{v}}_{\text{0}}}\] i.e. \[\frac{1}{2}m\,{{({{V}_{max}})}^{2}}=4\times \frac{1}{2}m\,{{(4\times {{10}^{6}})}^{2}}\] [from Eq. (i)] \[V_{\max }^{2}=4\times {{(4\times {{10}^{6}})}^{2}}\] \[{{V}_{\max }}=\sqrt{4\times {{(4\times {{10}^{6}})}^{2}}}\] \[=2\times 4\times {{10}^{6}}\] \[=8\times {{10}^{6}}\text{m}{{\text{s}}^{-1}}\]
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