Light of energy 5.0 eV falls on a metal of work function 3.0 eV and electrons are emitted, determine the stopping potential. [electronic charge, e = 1.60 x 108ms-19]
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Correct Answer: Option B
Explanation:
\(V_8 \frac{hf - w_o}{e} = \frac{5.0 \times 1.60 \times 10^{-19} - 3.0 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0V\)
\(V_8 \frac{hf - w_o}{e} = \frac{5.0 \times 1.60 \times 10^{-19} - 3.0 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0V\)