Question

The energy of a particular color of green light is 3.70E-22 kJ.

The wavelength of this light is nanometers.


(109 nm = 1 m)

Answer 1

so i first converted kj to k...3.70e-22kj*(1/1j)=3.70e-22k.....then.....3.70e-22k*(1000m/1k)=3.7e-19m...then....3.7e-19m*(10^9nm/1m)=3.7e-10

Answer 2

what did i do wrong???

Answer 3

I'm not sure I follow the conversions you are trying to do here.
kJ is kiloJoules, where kilo is just a prefix. Converting from kJ to k does not seem to make sense. We ought to convert from kJ to J and then apply a formula between Eenrgy and wavelength to solve for wavelength. Then convert wavelength from meters to nanometers...

Answer 4

oh so i have to use E=hv

Answer 5

Yes. Although v is frequency I believe, so we need just one more: c = (Lambda) v (c the speed of light, Lambda is wavelength, and v is frequency. This lets us convert between frequency and wavelength.
Or the cumulative formula is just E = (h c)/ (Lambda)

Answer 6

\( c = \lambda v \) \(E = h v = \dfrac{h c}{\lambda} \)

Answer 7

okay hold on im going to work it out

Answer 8

so i converted kj to j and i got 3.7e-19j would this be v or λ, we know that c=2.9979x10^8m/c

Answer 9

that would be energy: "The energy of a particular color of green light is..." So that goes in for E.
We want to find Wavelength, which is lambda. One way is by finding frequency v and then using the c = lambda * v formula to switch. Otherwise, we can just use this formula and find lambda directly.: E = hc / (lambda)

Answer 10

but we dont have λ

Answer 11

Lambda is what we want to find, yes? That is wavelength, which the problem is asking for.
h and c are just constants. And we have E given in the problem. Lambda is the only variable in the equation, so we can solve for it. :)

Answer 12

okay v=5.36e-7?

Answer 13

λ=5.59e14

Answer 14

is this right

Answer 15

Hm.. 5.36 * 10^(-7) looks like the value for lambda I got.

\( E = \dfrac{hc}{\lambda} \)
\( 3.7 10^{-19} = \dfrac{(6.626*10^{-34}) (2.9979*10^8)}{\lambda} \)

Answer 16

3.7 * 10^(-19) , not 3.710.

Answer 17

wait i used \[\frac{ hc }{ v }\]

Answer 18

If you solve here:
\( 3.7 * 10^{-19} = (6.626 * 10^{-34} ) \nu \)
\( \nu = \dfrac{3.7 *10^{-19}}{6.626 * 10^{-34}} \)
comes out to 5.58 * 10^(14).

I think you just reversed their roles.

Answer 19

thats how i got v=5.36e-7

Answer 20

Yes; you effectively already had lambda, just switch v for lambda really.
Then that would be wavelength in meters. Convert it to nanometers for the final answer. :)

Answer 21

so you did E=hv....so when they give you E=hv=hc/v (they are 2 differeny equations not have to be used together?

Answer 22

Hmm... The original equation that I am looking at is this:
\( E = h \nu \)
\(E\) is energy in Joules, \(h\) is Planck constant in Joule-seconds. \( \nu \) is frequency in cycles per second.
\( c = \lambda \nu \) \( \nu = \dfrac{c}{\lambda} \)
This separate equation relates wavelength \(\lambda\) (meters) and frequency \(\nu \) to the speed of light.
Substitution \( \nu = \dfrac{c}{\lambda} \) into the first equation, we have:
\( E = h \dfrac{c}{\lambda} \)

The usefulness of either equation is based on whether you have/need wavelength or frequency. In this case, we need wavelength, so we use the second equation. What I said previously is that technically, you *can* use the first one and find frequency, but you need to then convert to wavelength using the middle equation.

Answer 23

oh okay that makes much more sense...wow thank you so much!

Answer 24

Glad to help! :)

Were you able to find the answer? I am coming up with about 530 nm, which checks out according to the visible light spectrum.

Answer 25

im doing it hold on

Answer 26

λ=5.37e-7....i got 537

Answer 27

Yep, sounds good to me. :)

Answer 28

its rights!!! yess thank you... could i ask you another question in a couple of mins?

Answer 29

I may have to sleep very soon (it's late here, 12am). I'll see what I can do. :)

Answer 30

thank you...let me know then i could ask you tom

Answer 31

Is it a similar question to this one?

Answer 32

yup

Answer 33

Should be fine now then? That shouldn't take too long.

Answer 34

okay ill let you

Answer 35

what are the units for frequency?

Answer 36

A local FM radio station broadcasts at an energy of 6.52E-29 kJ/photon.

Calculate the frequency at which it is broadcasting.
Frequency = MHz

(1 MHz = 106 sec -1)

i got 3.03e-6 MHz

Answer 37

What was your process to obtain that answer?

Answer 38

i converted kj to j.....\[\frac{ 6.52e-29kj }{photon }*\frac{ 1000j }{ ?1kj }=6.52e-26 j/photon\] then \[6.52e-26=\frac{ 6.626e-34*2.9979e8 }{ v }\] then \[v=\frac{ 1.98e-25 }{ 6.52e-26 }\] ....v=3.036809816

Answer 39

with 3.036809816/10^6=3.03e-6MHz

Answer 40

I believe this is the same error from earlier... in that equation, v is in place of where wavelength is.
\( E = \dfrac{h c}{\color{blue}\lambda} \) This is the equation with wavelength.

\( E = h \color{red}{\nu} \) This is the equation with frequency.

Answer 41

okay i got 98

Answer 42

98 MHz looks good to me now.

It seems once you have the correct equation, you do fine with the math work. :)

Answer 43

thank you so much for explaining it to me i was so confused before...

Answer 44

Yep, glad to help again. :D

I will be heading off to bed now if that is all. Best of luck and gnight!

Answer 45

good night...hope to talk to you again!