The graph between the stopping potential and frequency for two different metal plates P and Q are shown in the figure. Which of the metals has a greater threshold wavelength and work function?

To determine which of the metals, P or Q, has a greater threshold wavelength and work function, we need to analyze the graph of stopping potential vs. frequency for both metals. The stopping potential is related to the energy of the ejected photoelectrons, and from that, we can infer information about the work function and threshold wavelength.

Here's what you can infer from the graph:

Work Function (Φ): The work function of a metal is the minimum energy required to release an electron from the metal's surface. In the context of the photoelectric effect, it corresponds to the y-intercept of the stopping potential vs. frequency graph. The metal with the greater y-intercept has a greater work function.

Threshold Wavelength (λ₀): The threshold wavelength is related to the work function by the equation: λ₀ = hc / Φ, where h is Planck's constant and c is the speed of light. Since work function is inversely proportional to threshold wavelength, the metal with the higher work function will have a smaller threshold wavelength.

So, look at the graph, and find which metal plate, P or Q, has the higher y-intercept (stopping potential at zero frequency). The metal with the higher stopping potential (more negative) has the greater work function, and consequently, it will have a smaller threshold wavelength.

In summary:

Metal with the higher y-intercept has the greater work function.
The metal with the greater work function will have a smaller threshold wavelength.
Without the actual graph data, I can't provide a definitive answer, but you can analyze the graph using the above principles to determine which metal has the greater threshold wavelength and work function.