A) charge of the electron
B) work function of the metal
C) Planck's constant
D) ratio of the Planck's constant to electronic charge
Correct Answer: C
Solution :
If \[{{V}_{0}}\] is stopping potential, then \[e{{V}_{0}}=hv-h{{v}_{0}}\] But \[\frac{1}{2}mv_{\max }^{2}=e{{V}_{0}}\] \[\therefore \] \[\frac{1}{2}mv_{\max }^{2}=hv-h{{v}_{0}}\] or \[KE=hv-h{{v}_{0}}\] Comparing with \[y=mx+c,\] we get Slope of curve m = h (Planck's constant)
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