Answer:
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
Answer is:
18 electrons.
<span>The principal quantum number (n) is one
of four quantum numbers which are assigned to each electron in an atom to
describe that electron's state.
The azimuthal quantum number (l) is a quantum number for
an atomic orbital that determines its orbital angular
momentum and describes the shape of the orbital, l = 0....n-1.</span>
<span>For n = 3, l = 0, 1, 2.</span>
<span>l = 0, s orbital with 2 electron.</span>
<span>l = 1, p orbitals with 6 electrons.</span>
<span>l = 2, d orbitals with 10 electrons.</span>
<span>2 + 6 + 10 = 18 electrons.</span>
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