## Exercise 1.9 Sig Fig

$c=\lambda v$

Vincent Chiang 1L
Posts: 54
Joined: Fri Sep 29, 2017 7:05 am

### Exercise 1.9 Sig Fig

Based on the SSM, for the second row in the table, why don't the number of sig figs match? 5.0E14 Hz is 2 sig figs, and 600 nm is 1 sig fig... Thanks in advance.

Thuy-Anh Bui 1I
Posts: 56
Joined: Sat Jul 22, 2017 3:00 am
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### Re: Exercise 1.9 Sig Fig

I think that technically the answer would be 6.0 x 10^-7 m for wavelength. However, it is easier to identify and compare wavelengths if the value is given in nanometers so it would be converted to 600 nm. If you wanted to be specific about sig figs, you could write a bar above the first zero to indicate 2 sig figs, but for this exercise I don't think that it matters.

AnuPanneerselvam1H
Posts: 52
Joined: Fri Sep 29, 2017 7:07 am

### Re: Exercise 1.9 Sig Fig

I agree, technically the answer should be 6.0 x 10^-7 so that the the number of sig figs remains two. They wrote it as 600 nm so that we can see the wavelength is part of the visible light spectrum.

William Cryer 1L
Posts: 12
Joined: Fri Sep 29, 2017 7:05 am

### Re: Exercise 1.9 Sig Fig

I also believe that they want us to convert the wavelengths from nm to m, it's another way to practice. For example, you would have to do extra dimensional analysis to solve for energy, wavelength, and frequency since they are not the same units. Just a way to challenge our problem solving skills, I think.