The maximum energy of photoelectrons emitted from potassium is \(2.1 eV\) when illuminated by light of wavelength \(3 \times 10^{-7}\) m and \(0.5 eV\) when the light wavelength is \(5\times 10^{-7}\) m. Use these results to obtain values for the minimum energy needed to free an electron from potassium, in \(eV\).

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