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When light of wavelength 350 nm falls on a potassium surface, electrons having a maximum kinetic energy of 1.31 eV are emitted. Find the cutoff wavelength.

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Given values:

λ=350 nm=350109 m\lambda=350 \ \text{nm}=350 \cdot 10^{-9} \ \text{m} KE=1.31 eVKE=1.31 \ \text{eV} ϕ=2.24 eV\phi=2.24 \ \text{eV}

In part b)\textbf{b)} we have to find the cutoff wavelength.

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