When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy \(T_A\) eV and de-Broglie wavelength \(\lambda_A\). The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is \(T_B=(T_A-1.5)\) eV. If the de-Broglie wavelength of these photoelectrons \(\lambda_B=2\lambda_A\), then the work function of metal B is :
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- 1.5 eV
- 4 eV
- 2 eV
- 3 eV
The Correct Option is B
Solution and Explanation
\(\dfrac{\lambda_B}{\lambda_A}=2\Rightarrow\sqrt{\dfrac{T_A}{T_B}}=2\Rightarrow T_A=4T_B\).
Given \(T_B=T_A-1.5\Rightarrow T_B=4T_B-1.5\Rightarrow 3T_B=1.5\Rightarrow T_B=0.5\) eV (and \(T_A=2\) eV).
Einstein's equation for B: \(\phi_B=h\nu-T_B=4.50-0.5=4.0\) eV.
Answer: (2) 4 eV
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