Monday

July 28, 2014

July 28, 2014

Posted by **Sarah** on Monday, October 26, 2009 at 4:01pm.

do I use E = -RH / n^2 ?

I don't know why I'm given the cm to J conversion, or why they ask for the wavelength first.

- Chemistry -
**DrBob222**, Monday, October 26, 2009 at 5:08pmI would do it the other way around.

delta E = 2.180 x 10^-18 J*[(1/n1^2)-(1/n2^2)]. With the Balmer series, n1 is 2 and n2 is infinity.

Then delta E = hc/wavelength.

The reason reciprocal cm is given (cm^-1) is because wave number = 1/wavelength and many spectroscopists prefer to use wave number instead of wavelength. Also, if you use the Rydberg constant, then

1/wavelength = R[(1/N1^2) - (1/N2^2)] and you don't need to convert to wavelength first to get energy in joules.

- Chemistry -
**Sarah**, Monday, October 26, 2009 at 6:01pmok so 1/lambda = (2.18e-19) (.25 - 0)

1/lambda = 5.45e-19

then i set up a proportion to convert it into joules?

1 cm-1/1.986e-23 joules = 5.45e-19 cm-1/x

I got x = 1.082e-41 joules and that it so not right..

- Chemistry -
**Sarah**, Monday, October 26, 2009 at 6:02pmok so 1/lambda = (2.18e-19) (.25 - 0)

1/lambda = 5.45e-19

then i set up a proportion to convert it into joules?

1 cm-1/1.986e-23 joules = 5.45e-19 cm-1/x

I got x = 1.082e-41 joules and that it so not right..

- Chemistry -
**Sarah**, Monday, October 26, 2009 at 9:12pmthank you i got it now!!

thanks so much for your help!!

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