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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

  1. A \(2 \times 10^{16}\)
  2. B \(4 \times 10^{16}\)
  3. C \(2.5 \times 10^{16}\)
  4. D \(1.33 \times 10^{16}\)
Verified Solution

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…