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(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$. Using $hc \approx 1240 \ eV \cdot nm$:For $\lambda_1 = 310 \ nm$,$E_1 = \frac{1240}{3

Vedclass · Accepted Answer

$2.8$

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(B) The energy of a photon is given by $E = \frac{hc}{\lambda}$. Using $hc \approx 1240 \ eV \cdot nm$:For $\lambda_1 = 310 \ nm$,$E_1 = \frac{1240}{310} = 4.0 \ eV$.For $\lambda_2 = 455 \ nm$,$E_2 = \frac{1240}{455} \approx 2.72 \ eV$.For $\lambda_3 = 620 \ nm$,$E_3 = \frac{1240}{620} = 2.0 \ eV$.According to Einstein's photoelectric equation,$K_{max} = E - \Phi$,where $\Phi = 1.2 \ eV$.For $\lambda_1 = 310 \ nm$,$K_{max} = 4.0 - 1.2 = 2.8 \ eV$.For $\lambda_2 = 455 \ nm$,$K_{max} = 2.72 - 1.2 = 1.52 \ eV$.For $\lambda_3 = 620 \ nm$,$K_{max} = 2.0 - 1.2 = 0.8 \ eV$.The maximum kinetic energy is the highest value among these,which is $2.8 \ eV$.

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