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We have Threshold wavelength $\lambda_0=680×10^{-9}m$ , and wavelength of incident radiation $\lambda=560×10^{-9}m$ By Einstein's equation we know that, $E=W+T_{max}\\ T_{max}=E-W$

Where $E$ is the energy of the incident photon, $W$ is the photoelectric work function and $T_{max}$ is the maximum kinettic energy. But we also have $E=h\delta=\dfrac c\lambda$ and $W=h\delta_0=\dfrac c{\lambda_0}$ .

Hence the equation becomes, $T_{max}=\dfrac{hc}\lambda-\dfrac{hc}{\lambda_0}\\=hc\left(\dfrac 1\lambda-\dfrac 1{\lambda_0}\right)\\=6.625×10^{-34}×3×10^8\left(\dfrac 1{560×10^{-9}}-\dfrac 1{680×10^{-9}}\right)\\T_{max}=\boxed{6.26×10^{-20}J}$

And the solution is $6.26×10{-20}×10^{20}=\boxed{6.26}$