A) \[e\]
B) \[h\]
C) \[\frac{h}{e}\]
D) \[he\]
Correct Answer: C
Solution :
Relation between threshold frequency - \[({{v}_{0}})\] and potential \[{{V}_{0}}\] is \[e{{V}_{0}}=h{{v}_{0}}-\phi \] So, \[{{V}_{0}}=\frac{h}{e}({{v}_{0}})-\frac{\phi }{e}\] Hence, slope of the graph is \[\frac{h}{e}\]
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