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When photons of energy 4.28 \mathrm{eV} strike the surface of a metal, the ejected photoelectrons have a maximum kinetic energy \mathrm{E}_{\mathrm{A}} \ \mathrm{eV} and de-Broglie wavelength \lambda_{\mathrm{A}}. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 \mathrm{eV} is E_B=(E_A-1.5) \mathrm{eV}. If The de-Broglie wavelength of these photoelectrons is  \lambda_{\mathrm{B}}=2 \lambda_{\mathrm{A}}, then
 

Option: 1

 the work function of \mathrm{A} is 2.25 \mathrm{eV}

 


Option: 2

 the work function of \mathrm{B}  is 4.28 \mathrm{eV}


Option: 3

\mathrm{E}_{\mathrm{A}}=2.0 \mathrm{eV}


Option: 4

\mathrm{E}_{\mathrm{B}}=2.75 \mathrm{eV}


Answers (1)

best_answer

\begin{aligned} & \text { As, } \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} ; \text { so } \frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=\sqrt{\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{E}_{\mathrm{B}}}} \\ & \text { or } 2=\sqrt{\frac{E_A}{E_B}} \quad \text { or } \quad E_A=4 E_B \\ & \text { or } \quad \frac{E_A}{4}=E_A-1.5 \quad \text { or } \quad E_A=2.0 \mathrm{eV} \\ & \phi_{\mathrm{A}}=4.28-2.00=2.28 \mathrm{eV} \\ & \phi_B=4.70-0.50=4.20 \mathrm{eV} \\ & \end{aligned}

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