AP EAMCET · PHYSICS · Dual Nature of Matter
Energy required to remove an electron from aluminum surface is 4.2 eV . If light of wavelength \(2000 Å\) falls on the surface, the velocity of the fastest electron ejected from the surface will be
- A \(8.4 \times 10^5 \mathrm{~ms}^{-1}\)
- B \(7.4 \times 10^5 \mathrm{~ms}^{-1}\)
- C \(6.4 \times 10^5 \mathrm{~ms}^{-1}\)
- D \(8.4 \times 10^6 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(A) \(8.4 \times 10^5 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\phi_{\mathrm{o}}=4.2 \mathrm{eV}, \lambda=2000 Å, \mathrm{v}_{\max }=?\) \(\therefore \mathrm{E}=\frac{12400}{\lambda(\mathrm{in} Å)} \mathrm{eV}=\frac{12400}{2000}=6.2 \mathrm{eV}\) By Einstein's photoelectric equation,…
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