Which of the following statements are correct, if the threshold frequency of caesium is 5.16 times 10^14 mathrm~Hz$5.16 \times 10^{14} \mathrm{~Hz}$ ?

A. When mathrmCs$\mathrm{Cs}$ is placed inside a vacuum chamber with an ammeter connected to it and yellow light is focused on mathrmCs$\mathrm{Cs}$ the ammeter shows the presence of current. B. When the brightness of the yellow light is dimmed, the value of the current in the ammeter is reduced. C. When a red light is used instead of the yellow light, the current produced is higher with respect to the yellow light. D. When a blue light is used, the ammeter shows the formation of current. E. When a white light is used, the ammeter shows formation of current. Choose the correct answer from the options given below:

### Related Formula lambda = fraccnu$$\lambda = \frac{c}{\nu}$$ where, c = 3.0 times 10^8 text m s^-1$c = 3.0 \times 10^8 \text{ m s}^{-1}$ nu_0 = 5.16 times 10^14 text Hz$\nu_0 = 5.16 \times 10^{14} \text{ Hz}$ ### Core Logic Let's first calculate the threshold wavelength (lambda_0$\lambda_0$) for Caesium: lambda_0 = frac3.0 times 10^85.16 times 10^14 approx 5.81 times 10^-7 text m = 581.4 text nm$$\lambda_0 = \frac{3.0 \times 10^8}{5.16 \times 10^{14}} \approx 5.81 \times 10^{-7} \text{ m} = 581.4 \text{ nm}$$ Comparing this with the visible spectrum: - Yellow light (sim 580 text nm$\sim 580 \text{ nm}$) corresponds closely to the threshold wavelength. Hence, photoelectric emission occurs (Statement A is correct). - The photocurrent is directly proportional to the light intensity (brightness). Reducing brightness decreases the current (Statement B is correct). - Red light (sim 620text--750 text nm$\sim 620\text{--}750 \text{ nm}$) has a lower frequency than the threshold frequency, meaning no current is produced (Statement C is incorrect). - Blue light (sim 450text--490 text nm$\sim 450\text{--}490 \text{ nm}$) has a higher frequency than the threshold. Thus, current is produced (Statement D is correct). - White light contains all wavelengths (including blue, violet, etc.), which will produce a photoelectric current (Statement E is correct). Therefore, statements A, B, D, and E are correct. ### Pattern Recognition The wavelength threshold is 581 text nm$581 \text{ nm}$ (yellow-green region). Any light with shorter wavelength (higher frequency), like blue or white, will cause emission. Dimming simply decreases photon flux and hence current. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom Class 12 Physics: Dual Nature of Radiation and Matter