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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \Phi$

Vedclass · Accepted Answer

$10$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = E - \Phi$,where $E$ is the energy of the incident photon and $\Phi$ is the work function of the material.Given $E = 4.5 \ eV$ and $\Phi = 3 \ eV$,we have $K_{max} = 4.5 \ eV - 3 \ eV = 1.5 \ eV$.Converting this energy to Joules: $K_{max} = 1.5 	imes 1.6 	imes 10^{-19} \ J = 2.4 	imes 10^{-19} \ J$.The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mK_{max}}}$,where $h = 6.63 	imes 10^{-34} \ J \cdot s$ and $m = 9.1 	imes 10^{-31} \ kg$.Substituting the values: $\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 2.4 	imes 10^{-19}}} = \frac{6.63 	imes 10^{-34}}{\sqrt{43.68 	imes 10^{-50}}} = \frac{6.63 	imes 10^{-34}}{6.61 	imes 10^{-25}} \approx 1.0 	imes 10^{-9} \ m$.Since $1 \ Å = 10^{-10} \ m$,we have $\lambda \approx 10 \ Å$.

Photons of energy $4.5 \ eV$ are incident on a photosensitive material of work function $3 \ eV$. The de Broglie wavelength associated with the photoelectrons emitted with maximum kinetic energy is nearly (in $Å$)

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