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Chemistry 12 Online
OpenStudy (anonymous):

What is the wavelength (in nm) of radiation that has an energy content of 2.13x10^3 kJ/mol? In which region of the electromagnetic spectrum is this radiation found? Can someone explain me this step by step? Thanks

OpenStudy (kira_yamato):

\[E = \frac{hc}{\lambda}\] Plug in E = 2.13 x 10^3

OpenStudy (kira_yamato):

Make sure you convert your units correctly.

OpenStudy (anonymous):

I was checking my equations and I have none with the hc together :o

OpenStudy (kira_yamato):

Do you have this equation? \[E = h \nu\]

OpenStudy (anonymous):

E=hv yes

OpenStudy (kira_yamato):

\[\nu = \frac{c}{\lambda}\] right? Plug this into that equation (:

OpenStudy (anonymous):

that's another equation =.=

OpenStudy (kira_yamato):

It's like solving simultaneous equations \[\nu = \frac{c}{\lambda}\] ----(1) \[E = h \nu\] ----(2) Plug (1) into (2) \[E = \frac{hc}{\lambda}\]

OpenStudy (anonymous):

2.13x10^3 = 2.99792458x10^8m/s / wavelength

OpenStudy (kira_yamato):

Where's your h?

OpenStudy (anonymous):

uh oh then I have to use E=hc/λ

OpenStudy (anonymous):

what would h be then?

OpenStudy (anonymous):

I am given the Energy of the content

OpenStudy (anonymous):

+ I know the velocity of light

OpenStudy (anonymous):

Wouldn't I have to use E=hv instead?

OpenStudy (anonymous):

oh nvm.. I need the wavelength but what is h then D:

OpenStudy (kira_yamato):

h = 6.63 x 10^–34 J s

OpenStudy (anonymous):

Plank's .. got it

OpenStudy (anonymous):

after the calculation, how do I know in which region of the spectrum is this located?

OpenStudy (anonymous):

2.13x10^3 = (6.63x10^-34)(2.99792458x10^8) / wavelength

OpenStudy (anonymous):

wavelength = 1071631256066461335740540575.9149 or 1.07x10^27 hertz ?

OpenStudy (anonymous):

I feel like I'm doing it wrong ... the wavelength is too long! lol

OpenStudy (anonymous):

|dw:1383100793497:dw|

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