Find the maximum magnitude of the linear mome | vedclass

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{1240 \, eV \cdot nm}{400 \, nm} = 3.1 \, eV$.According to Einstein's

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$4 	imes 10^{-25} \, kg \cdot m/s$

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{1240 \, eV \cdot nm}{400 \, nm} = 3.1 \, eV$.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is $K_{max} = E - \phi$,where $\phi = 2.5 \, eV$.$K_{max} = 3.1 \, eV - 2.5 \, eV = 0.6 \, eV$.Converting $K_{max}$ to Joules: $K_{max} = 0.6 	imes 1.6 	imes 10^{-19} \, J = 0.96 	imes 10^{-19} \, J$.The relationship between momentum $p$ and kinetic energy $K$ is $p = \sqrt{2mK}$.$p = \sqrt{2 	imes (9.1 	imes 10^{-31} \, kg) 	imes (0.96 	imes 10^{-19} \, J)}$.$p = \sqrt{17.472 	imes 10^{-50}} \approx 4.18 	imes 10^{-25} \, kg \cdot m/s$.Rounding to the nearest option,the value is approximately $4 	imes 10^{-25} \, kg \cdot m/s$.

Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength $400 \, nm$ falls on a metal having a work function of $2.5 \, eV$.

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