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What..can comeone help me with my chemistry?

Please help me with these problems, I have no clue how to do I google it and some one had a similar question and I didn’t understand what he was talking about, so um help please

  1. Determine the frequency of light whose wavelength is 4.357 x 10^ -7 cm.

  2. Determine the energy in joules of a photon whose frequency is 3.55 x 10^ 17 hz.

13.how long would it take a radio wave whose frequency is 7.25 x 10^5 hz to travel from mars to earth if the distance between the two planets is approximately 8.00 x 10^7 km?

  1. Using the equations e=hv and c =λv, derive an equation expressing e in terms of h, c, and λ.

  2. Cobalt is an artificial radioisotope that is produced in a nuclear ray source in the treatment of certain types of cancer. If the wavelength of the gamma radiation from a cobalt -60 source is 100 x 10^ -3 nm, calculate the energy of a photon of this radiation.

Know its a lot but I really don’t have a clue to do this and thank you so much to whomever answers it correctly.

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Answer #1

This is physics dude…not chemistry. Anyway here are the answers.

(11) The wavelength is 4.357 10^(-7) cm –> 4.357 10(-9) m. Using the fact that the speed of light = 3 * 10^8 m/s and c =λf

therefore f = c/λ

               =  [ 3 * 10^8 ] / [4.357 * 10^(-9) ]
               = 6.885 * 10^16 Hz

(12) Using the equation e = hf e = (6.36 10^(-34)) (3.55 * 10^17) = 2.258 J

(13) d = vt therefore , t = d/v

               t = (8.00 * 10^7) / (3 * 10^8)
                 = 0.267 seconds

(14) e = hf …(1) c =λf …(2)

Make the f the subject of (2) f = c/λ …(3)

Sub (3) into (1)

e = hf = h(c/λ) = hc/λ

(15) e = hc/λ = 6.36 * 10^(-34) / (100 10^(-12)) = 1.908 10^(-15) J

Answer #2

Sorry but I used the wrong value for Planck’s Constant, h. I used h = 6.36 10^(-34) whereas it should be h = 6.63 10^(-34)

So I’ll rewrite the correct solutions for you

(12) e =2.354 10^(-16) J (15) e = 1.989 10^(-15) J

Sorry for the mistake.

Answer #3

thanks ^_^ and I don’t know why were doing physics in chemistry then? but thank you so much

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