A metal was irriadated by light of frequency `3.2xx10^(15)S^(-1)`. The photoelectron prouduced had its KE, 2 times the KE of the photoelectron which was product when the same metal was irriadated with a light of frequency `2.0xx10^(15)S^(-1)`. What is work function.

Explanation

To solve the problem, we need to use the photoelectric effect equation, which relates the energy of the incoming photons to the kinetic energy of the emitted photoelectrons and the work function of the metal. The equation is given by:

\[ E = KE + \phi \]

Where:
- \( E \) is the energy of the incoming photon,
- \( KE \) is the kinetic energy of the emitted photoelectron,
- \( \phi \) is the work function of the metal.
...