1. What is the frequency of light with a wavelength of 1064 nm? 2. How many photons of 1064 nm light are contained in a 535 Joules of this light?

1) 2.8×10^14 Hz

2) 2.88×10^21 photons

Explanation:

Recall that the formula for the speed of a wave is v=λf, since we are talking about light, we can replace v with c hence; c=λf. Where;

c= speed of light = 3 ×10^8 m/s

λ= wavelength of light= 1064nm= 1064×10^-9 m

f= frequency of light= the unknown

Hence;

f= c/λ= 3×10^8/1064×10^-9

f= 2.8×10^14 Hz

Energy of a single photon=hf= 6.626×10^-34 Js × 2.8×10^14 s^-1 = 18.55×10^-20J

If 1 photon contains 18.55×10^-20J of energy

x photons contains 535 J of energy

x= 535/18.55×10^-20J

x= 2.88×10^21 photons

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.