A) \[E=m{{c}^{2}}\]
B) \[E=hv~\]
C) \[hv=\frac{1}{2}m{{v}^{2}}\]
D) \[E=\frac{h}{\lambda }\]
Correct Answer: D
Solution :
In 1905, Einstein realized that the photoelectric effect could be understood if the energy in light is not spread out over wave fronts but is concentrated in small packets, or photons. Each photon of light of frequency v has the energy hv. Thus, Einstein's work on photoelectric effect gives support to \[E=hv\].
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