Answer: Option (b) is the correct.
Explanation:
It is known that energy is directly proportional to frequency.
Mathematically, Â Â Â Â Â Â Â [tex]E \propto \nu[/tex]
or, Â Â Â Â Â Â Â Â Â Â Â Â Â E = [tex]h \times \nu[/tex] ............ (1)
where, Â Â E = energy
        h = Plank's constant
       [tex]\nu[/tex] = frequency
This relation shows that higher is the energy then more will be the frequency of electrmagnetic radiation.
Also, Â Â [tex]\nu = \frac{c}{\lambda}[/tex] ......... (2)
where, Â Â Â Â Â c = speed of light
         [tex]\lambda[/tex] = wavelength
Therefore, substituting value of [tex]\nu[/tex] from equation (2) into equation (1) as follows.
         E = [tex]h \times \nu[/tex]
         E = [tex]h \times \frac{c}{\lambda}[/tex]
This relation shows that higher is the energy then lower will be the wavelength.
Thus, we can conclude that as the energy of a photon increases, its frequency decreases, is incorrect statement.