A) 6.82e V
B) 12.5e V
C) \[6.25\times {{10}^{5}}\]
D) \[4.87\times {{10}^{5}}\]
Correct Answer: D
Solution :
\[K{{E}_{Max}}=h{{v}_{Max}}-\phi \] |
\[=\frac{(6.6\times {{10}^{-34}})\,(6.28\times {{10}^{7}})(3\times {{10}^{8}})}{1.6\times {{10}^{-19}}\times 2\times 3.14}-4.7\] |
\[=12.37-4.7\]\[=7.67\text{ }eV.\] |
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