Statement I: Ultraviolet (UV) rays have the highest frequency among the given options (microwaves, infrared rays, and ultraviolet rays). Since the energy of a photon is directly proportional to its frequency (\( E = h f \)), UV rays are the most effective for the emission of electrons from a metallic surface. Hence, Statement I is true.
Statement II: The maximum kinetic energy of photoelectrons is given by the photoelectric equation: \[ K.E_{\text{max}} = h f - h f_0, \] where \( h \) is Planck's constant, \( f \) is the frequency of the incident light, and \( f_0 \) is the threshold frequency. From the equation, it is clear that \( K.E_{\text{max}} \) is directly proportional to \( f \), not inversely proportional.
Hence, Statement II is false. Thus, the correct answer is \( \boxed{(3)} \).
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: