What is the approximate frequency of a photon having an energy 3.2 * 10^-24 j? (h = 6.6*10^-34 j*s) a.) 48*10^9 b.) 2.1*10^-57 c.) 3.0*10^-58 d.) 2.1*10^-10

The Planck’s Equation states that

E = hv

E stands for energy measured in Joules

v stands for frequency measured in s^-1 or Hertz (Hz)- 1Hz .

h is Planck’s constant which is equivalent to 6.626x10-34 J.s

The Planck's equation states that the energy of a photon, is directly proportional to its frequency (v), by a constant factor (h).

Given:

E=3.2 * 10^-24 J

h= 6.626x10-34 J.s

Now we know that

E=hv

Thus substituting the given values we get

v=3.2 * 10^-24 /6.626x10-34

v=48 x 10^9 s^-1

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

level 4 to level 2 : ))