Violet light has a wavelength of 4.50 x 10-12 m. What is the frequency?* 3 points

1.5 x10^20 1/s

0.0135 1/s

6.77x10^19 1/5

3x10^8 1/s

Green light has a frequency of 6.73 x 1014 1/s. What is the wavelength?*

3 points

2.24x10^6 m

4.46x10^-7 m

2.02 x 10^23 m.

4.95 x 10^-24 m

3 points

What is the energy (Joules) of violet light with a frequency = 3.43 x 10^14

1/s? *

4.54x10^-19 J

1.99x10^-25 J

1.93 x 10^-48 J

2.27 x 10^-19 J

Frequency = 6.67x10¹⁹ s⁻¹

Wavelength = 4.46x10⁻⁷m

e = 2.27x10⁻¹⁹J

Explanation:

To convert frequency to wavelength and vice versa we use the equation:

Wavelength = Speed of light / Frequency

Speed ligth is 3x10⁸m/s

For a wavelength of 4.50x10⁻¹²m:

4.50x10⁻¹²m = 3x10⁸m/s / Frequency

Frequency = 3x10⁸m/s / 4.5x10⁻¹²m

Frequency = 6.67x10¹⁹ s⁻¹

For a frequency of 6.73x10¹⁴s⁻¹:

Wavelength = 3x10⁸m/s / 6.73x10¹⁴s⁻¹

Wavelength = 4.46x10⁻⁷m

And energy, e, from frequency, is obtained as follows:

e = h ₓ frequency

Where h is Planck's constant, 6.626x10⁻³⁴J*s

e = 6.626x10⁻³⁴J*s*3.43x10¹⁴s⁻¹

e = 2.27x10⁻¹⁹J

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.