TS EAMCET · Physics · Dual Nature of Matter
The work function of a metal is \(2 \mathrm{eV}\). If a radiation of wavelength \(3000 Å\) is incident on it, the maximum kinetic energy of the emitted photoelectrons is (Planck's constant \(h=6.6 \times 10^{-34} \mathrm{Js}\); velocity of light \(\left.c=3 \times 10^8 \mathrm{~m} / \mathrm{s} ; 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)\)
- A \(4.4 \times 10^{-19} \mathrm{~J}\)
- B \(5.6 \times 10^{-19} \mathrm{~J}\)
- C \(3.4 \times 10^{-19} \mathrm{~J}\)
- D \(2.5 \times 10^{-19} \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(3.4 \times 10^{-19} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\[ \text { Maximum } \mathrm{KE}=\frac{h \mathrm{c}}{\lambda}-\phi_0 \] Given, \(\lambda=3000 Å=3000 \times 10^{-10} \mathrm{~m}\), \[ h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}, \mathrm{C}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \phi=2 \mathrm{eV} \] Maximum KE…
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