What is the energy of a photon of red light having a frequency of 3.08 x 10^14 HZ?

2.04 x 10⁻¹⁹J

Explanation:

Given parameters:

  Frequency = 3.08 x 10¹⁴Hz

Unknown:

The energy of the photon  = ?

Solution:

 The energy of a photon is given as;

          E = hf

Where h is the planck's constant

            f is the frequency

Insert the parameters and solve;

           E = 6.63 x 10⁻³⁴ x 3.08 x 10¹⁴

          E = 20.4 x 10⁻³⁴⁺¹⁴

           E = 20.4 x 10⁻²⁰  = 2.04 x 10⁻¹⁹J

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.