In a photoelectric effect experiment,the cath | vedclass

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$

Vedclass · Accepted Answer

$1.02$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For $\lambda_1 = 600 \ nm$,$K_1 = \frac{hc}{\lambda_1} - \phi$  --- $(i)$For $\lambda_2 = 400 \ nm$,$K_2 = 2K_1 = \frac{hc}{\lambda_2} - \phi$ --- (ii)From $(i)$,$K_1 = \frac{hc}{\lambda_1} - \phi$. Substituting this into (ii):$2 \left( \frac{hc}{\lambda_1} - \phi ight) = \frac{hc}{\lambda_2} - \phi$$\frac{2hc}{\lambda_1} - 2\phi = \frac{hc}{\lambda_2} - \phi$$\phi = hc \left( \frac{2}{\lambda_1} - \frac{1}{\lambda_2} ight)$Substituting the values: $h = 6.63 	imes 10^{-34} \ J-s$,$c = 3 	imes 10^8 \ m/s$,$\lambda_1 = 600 	imes 10^{-9} \ m$,$\lambda_2 = 400 	imes 10^{-9} \ m$:$\phi = (6.63 	imes 10^{-34} 	imes 3 	imes 10^8) \left( \frac{2}{600 	imes 10^{-9}} - \frac{1}{400 	imes 10^{-9}} ight)$$\phi = (19.89 	imes 10^{-26}) 	imes 10^9 \left( \frac{1}{300} - \frac{1}{400} ight)$$\phi = 19.89 	imes 10^{-17} \left( \frac{4-3}{1200} ight) = \frac{19.89 	imes 10^{-17}}{1200} \ J$$\phi = \frac{19.89 	imes 10^{-17}}{1200 	imes 1.6 	imes 10^{-19}} \ eV \approx 1.036 \ eV \approx 1.02 \ eV$.

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