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Chemistry 14 Online
OpenStudy (anonymous):

Plank's Constant

OpenStudy (anonymous):

@Abhisar

OpenStudy (anonymous):

http://prntscr.com/582t1f

OpenStudy (abhisar):

#a E=h\(\sf \nu\) => \(\sf 6.6 \times 10^{-34} \times 3.61 \times 10^{11}\) = \(\sf 23.826 \times 10^{-12}\) joules

OpenStudy (abhisar):

Got it ?

OpenStudy (anonymous):

Yup

OpenStudy (abhisar):

#b Similarly, \(\sf E=h\frac{c}{\lambda}\\ \sf c=speed~of~light\\ \sf \lambda =wavelength~in~metre\)

OpenStudy (abhisar):

Notice that the wavelength given in 2nd question is in Nano metres. Convert it into metre and plug in the values in the above formula and finally find the value of energy in joules.

OpenStudy (abhisar):

Got it ?

OpenStudy (anonymous):

So far so good

OpenStudy (abhisar):

ok, so you done ?

OpenStudy (anonymous):

What should Ive gotten for number 2?

OpenStudy (abhisar):

I havn't ca;cu;ated it...hang on

OpenStudy (anonymous):

Ok sure

OpenStudy (abhisar):

*calculated

OpenStudy (abhisar):

\(\approx 3.18 \times 10^{-19}\)

OpenStudy (abhisar):

What have you got ?

OpenStudy (anonymous):

Ok I was completely different

OpenStudy (anonymous):

check and you were right though

OpenStudy (anonymous):

so thats good

OpenStudy (abhisar):

\(\sf \huge \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.21 \times 10^2 \times 10^{-9}}\)

OpenStudy (anonymous):

first one was wrong though Im assuming I provided too many digits?

OpenStudy (abhisar):

No, i have done a mistake there......in first one it will be \(\sf 23.826 \times 10^{-23}\)

OpenStudy (anonymous):

Aw shoot

OpenStudy (abhisar):

or properly \(\sf 2.3826 \times 10^{-22}\)

OpenStudy (anonymous):

Ok thanks for your help!

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