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Chemistry 7 Online
OpenStudy (kittiwitti1):

Which of the following defines a correct relationship for electromagnetic radiation? a. \(\lambda\alpha v\) b. \(\lambda\alpha E\) c. \(E\alpha\frac{1}{v}\) d. \(v\approx c\) e. \(v\lambda=c\)

OpenStudy (kittiwitti1):

the \(\alpha\) is supposed to be this:|dw:1469040501613:dw|

OpenStudy (kittiwitti1):

@mayankdevnani

OpenStudy (mayankdevnani):

what do you think?

OpenStudy (kittiwitti1):

idk, does the \(\alpha\) symbol mean proportional?

OpenStudy (mayankdevnani):

yea

OpenStudy (kittiwitti1):

ok so uhh \(\lambda\)=wavelength and \(E\)=? I think \(v\) is velocity?

OpenStudy (mayankdevnani):

E=energy and v(mue) is frequency

OpenStudy (kittiwitti1):

mue? ok so a. wavelength \(\alpha\) energy b. wavelength \(\alpha\) energy c. energy \(\alpha\) \(\frac{1}{v}\) d. frequency \(\approx\) \(c\) e. frequency * wavelength = \(c\) also, what is \(c\)?

OpenStudy (mayankdevnani):

|dw:1469041448387:dw|

OpenStudy (mayankdevnani):

c is speed of light

OpenStudy (kittiwitti1):

ohh, okay o: so then \(\color{blue}{\text{Originally Posted}}\) wavelength \(\alpha\) energy b. wavelength \(\alpha\) energy c. energy \(\alpha\) \(\frac{1}{v}\) d. frequency \(\approx\) light speed e. frequency * wavelength = light speed \(\color{blue}{\text{End of Quote}}\)

OpenStudy (mayankdevnani):

what do you think? any guesses ?

OpenStudy (kittiwitti1):

umm D and E are wrong because light speed is a set #, idk anything else though

OpenStudy (mayankdevnani):

E is right it's a formula

OpenStudy (kittiwitti1):

oh... okay

OpenStudy (kittiwitti1):

hmm... I think A is right, idk about B, C is maybe

OpenStudy (mayankdevnani):

option A is wrong na check option E

OpenStudy (kittiwitti1):

but it is a real formula yes? xD

OpenStudy (mayankdevnani):

yea

OpenStudy (mayankdevnani):

so do you agree that option A is wrong ?

OpenStudy (kittiwitti1):

idk about relationship for electromag radiation lol

OpenStudy (kittiwitti1):

does the radiation have to involve light energy?

OpenStudy (mayankdevnani):

yo

OpenStudy (kittiwitti1):

ohhhh I see o:

OpenStudy (mayankdevnani):

so option A is incorrect

OpenStudy (kittiwitti1):

is not relevant to the question basically. o: ? that means b and c also

OpenStudy (kittiwitti1):

so then (this is from another question); this is also right?? \(\text{"The speed of light is equal to the product of the wavelength}\)\(\text{and frequency of an electromagnetic wave"}\)

OpenStudy (mayankdevnani):

yea

OpenStudy (mayankdevnani):

smart lady

OpenStudy (kittiwitti1):

hahah ty > <

OpenStudy (mayankdevnani):

*girl

OpenStudy (kittiwitti1):

aye, am 18, already lady :v

OpenStudy (kittiwitti1):

lol

OpenStudy (mayankdevnani):

lady nuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

OpenStudy (mayankdevnani):

;(

OpenStudy (kittiwitti1):

hehe

OpenStudy (mayankdevnani):

you're my girl <3

OpenStudy (kittiwitti1):

O_o

OpenStudy (mayankdevnani):

lets back on question,so what are your answers ?

OpenStudy (kittiwitti1):

I answer E o: also, this statement \(\text{Electromagnetic radiation travels at a finite speed}\) is it right?

OpenStudy (mayankdevnani):

E is correct

OpenStudy (mayankdevnani):

yea! correct

OpenStudy (kittiwitti1):

alright! thank you ☺ sorry, I got to go to chem class now ; - ;

OpenStudy (mayankdevnani):

cuz speed of light is a electromagnetic wave and its value is finite

OpenStudy (mayankdevnani):

☺ ☺ ☺ dun leave me alone

OpenStudy (mayankdevnani):

pm me

OpenStudy (kittiwitti1):

im sorry ; - ; OS not for android ; - ;

OpenStudy (kittiwitti1):

byebye </3

OpenStudy (mayankdevnani):

OS is for android as well use chrome ,you can only pm in mobile

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