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  • is the energy of incident photon and the work function is <img alt="2 \ \mathrm{e}V" src="/latex-image/?2%20%5C%20

4 \ \mathrm{e}V is the energy of incident photon and the work function is 2 \ \mathrm{e}V. The stopping potential will be
 

Option: 1

2V


Option: 2

4V


Option: 3

6V


Option: 4

2\sqrt{2}V


Answers (1)

best_answer

\begin{aligned} &\text{Einstein equation } \mathrm{E}=\mathrm{hv}_0+\mathrm{K} . \mathrm{E}.\\ &\text{Where }\mathrm{E}= \text{energy of incident photon }\\ &\mathrm{hv}_0= \text{work function of metal }\\ &\text{ K.E. = max. kinetic energy of } \mathrm{e}^{-}\\ &\therefore \quad 4 \mathrm{e}V=2 \mathrm{e}V+ \mathrm{K.E.} \text{ or }\mathrm{K.E.} =2 \mathrm{e}V \end{aligned}

Stopping potential is the potential difference that may stop this \mathrm{e}^{-}.
\text{Let it be }V, \text{then }\mathrm{e}V=2 \mathrm{e} \Rightarrow V=2 \text{volt}.

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