1. **Analysis of Assertion A:**
The intensity of light \( I \) can be expressed as:
\[ I = \frac{nh\nu}{A}, \] where \( n \) is the number of photons per unit time, \( h \) is Planck’s constant, \( \nu \) is the frequency, and \( A \) is the area. Rearranging for \( n \):
\[ n = \frac{IA}{h\nu}. \] For a constant intensity \( I \), if the frequency \( \nu \) increases, the number of photons \( n \) decreases. Thus, Assertion A is incorrect.
2. **Analysis of Reason R:**
According to the photoelectric effect, the maximum kinetic energy of emitted electrons is given by:
\[ K_{\text{max}} = h\nu - \phi, \] where \( \phi \) is the work function of the material. As frequency \( \nu \) increases, \( K_{\text{max}} \) also increases. Therefore, Reason R is correct.
Thus, the correct answer is option **(4): Assertion A is not correct, but Reason R is correct.**
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]