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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10

Vedclass · Accepted Answer

$1.3 	imes 10^{-9} \, m$

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{660 	imes 10^{-9}} = 3.01 	imes 10^{-19} \, J$.The work function $\Phi = 1.0 \, eV = 1.6 	imes 10^{-19} \, J$.The kinetic energy of the ejected electron is $KE = E - \Phi = 3.01 	imes 10^{-19} - 1.6 	imes 10^{-19} = 1.41 	imes 10^{-19} \, J$.The de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2m KE}}$.Substituting $m = 9.1 	imes 10^{-31} \, kg$ and $h = 6.63 	imes 10^{-34} \, J \cdot s$:$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 1.41 	imes 10^{-19}}} \approx 1.32 	imes 10^{-9} \, m$.

$A$ metal is irradiated with light of wavelength $660 \, nm$. Given that the work function of the metal is $1.0 \, eV$,the de Broglie wavelength of the ejected electron is close to

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