AP EAMCET · Chemistry · Structure of Atom
When a certain metal was irradiated with light of frequency \(4.0 \times 10^{16} \mathrm{~s}^{-1}\), the photoelectrons emitted had four times kinetic energy as the kinetic energy of photoelectrons emitted when the same metal was irradiated with light of frequency \(2.0 \times 10^{16} \mathrm{~s}^{-1}\). The threshold frequency \(\left(v_0\right)\) of the metal in \(\mathrm{s}^{-1}\) is
- A \(2 \times 10^{16}\)
- B \(4 \times 10^{16}\)
- C \(2.5 \times 10^{16}\)
- D \(1.33 \times 10^{16}\)
Answer & Solution
Correct Answer
(D) \(1.33 \times 10^{16}\)
Step-by-step Solution
Detailed explanation
Given, \[ \begin{aligned} v_1 & =4 \times 10^{16} \mathrm{sec}^{-1} \\ v_2 & =2 \times 10^{16} \mathrm{sec}^{-1} \\ \mathrm{KE}_1 & =4 \mathrm{KE}_2 \end{aligned} \] From equation of photoelectric effect, On putting the above calculated value in eq (ii), we get…
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