TS EAMCET · Physics · Dual Nature of Matter
Photons of energy \(2.4 \mathrm{eV}\) and wavelength \(\lambda\) fall on a metal plate and release photoelectrons with a maximum velocity \(v\). By decreasing \(\lambda\) by \(50 \%\), the maximum velocity of photoelectrons becomes \(3 v\). The work function of the material of the metal plate is
- A \(2.1 \mathrm{eV}\)
- B \(1.7 \mathrm{eV}\)
- C \(2.8 \mathrm{eV}\)
- D \(2.0 \mathrm{eV}\)
Answer & Solution
Correct Answer
(A) \(2.1 \mathrm{eV}\)
Step-by-step Solution
Detailed explanation
Let \(\phi_0=\) work-function of metal. Then, according to Einstein's photoelectric equation, \(K_{\max }=\frac{1}{2} m v^2=\frac{h c}{\lambda}-\phi_0\) When wavelength is reduced by \(50 \%\) i.e, \(\lambda^{\prime}=\frac{\lambda}{2}\), then maximum velocity of emitted…
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