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

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

12. 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?

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

15. 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.

3 answers

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ANSWER #1 of 3

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)](3 * 10^8) / (100 * 10^(-12))

= 1.908 * 10^(-15) J

ANSWER #2 of 3

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.

please help am desbrought
ANSWER #3 of 3

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

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