Ask your own question, for FREE!
Chemistry 6 Online
OpenStudy (anonymous):

Ultraviolet light emits a total of 2.5 × 10^–17 J of light at a wavelength of 9.8 × 10^–7 m. How many photons does this correspond to?

OpenStudy (aaronq):

find how much energy one photon of this wavelength contains: E=hc/lambda then divide the total energy by the energy of the one photon.

OpenStudy (anonymous):

6.626*10^-34*3.00*10^8/9.8*10^-7

OpenStudy (aaronq):

looks good

OpenStudy (anonymous):

2.5*10^-17/2.028*10^-19

OpenStudy (aaronq):

yep

OpenStudy (anonymous):

WELL I GOT 123 AND THE ANSWER I NEED 125 WHAT DID I DO WRONG

OpenStudy (aaronq):

try not rounding up the speed of light

OpenStudy (anonymous):

OK

OpenStudy (anonymous):

CAN YOU HELP ME WITH THIS ONE Which of the following statements is true? a) Light with energy lower than the binding energy of an electron will result in an electron being ejected 0but having no kinetic energy. b) Light with energy higher than the binding energy of an electron will result in multiple electrons being ejected from the surface. c) Light with energy higher than the binding energy of an electron will result in an electron being ejected with the excess energy as kinetic energy.

OpenStudy (aaronq):

well which one makes sense?

OpenStudy (anonymous):

C

OpenStudy (aaronq):

yep, thats right. If a photon of excess energy to be able to ionize an electron removes it, the rest will be translated into kinetic energy and the electron will go flying off

OpenStudy (anonymous):

AND THIS ONE I NOT SURE I THINK ITS D Which of the following is true regarding electromagnetic radiation? a) As energy increases, wavelength and frequency increase. b) As energy increases, wavelength and frequency decrease. c) As energy increases, wavelength increases and frequency decreases. d) As energy increases, wavelength decreases and frequency increases.

OpenStudy (aaronq):

yep, you can verify by looking at plancks equation E=hv=hc/lambda

OpenStudy (anonymous):

THANK YOU SO MUCH

OpenStudy (aaronq):

no problem dude

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!