HW 1.A #9


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Doris Cho 1D
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Joined: Thu Jul 11, 2019 12:16 am

HW 1.A #9

Postby Doris Cho 1D » Thu Oct 10, 2019 10:14 pm

I matched up the right radiation to each event but I only figured out the 3.3 x 10^-19 J to microwaves because of process of elimination... does it need to be converted to Hz or nm for me to see that it goes with microwaves?

Jocelyn Thorp 1A
Posts: 103
Joined: Wed Sep 18, 2019 12:20 am

Re: HW 1.A #9

Postby Jocelyn Thorp 1A » Thu Oct 10, 2019 11:50 pm

Doris Cho 1D wrote:I matched up the right radiation to each event but I only figured out the 3.3 x 10^-19 J to microwaves because of process of elimination... does it need to be converted to Hz or nm for me to see that it goes with microwaves?


I've seen charts with Hz or m but never any with Joules, so that might be easier

alicechien_4F
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Joined: Sat Jul 20, 2019 12:15 am

Re: HW 1.A #9

Postby alicechien_4F » Thu Oct 10, 2019 11:53 pm

Hi Doris! I think the easiest way to match the wave to the activity is by using Hertz since most EM spectrum charts label the radiation based on hertz. Also, the answer key actually matches 3.3E-19 J (which has a frequency of 5.0E14 Hz) to reading instead of microwave, which makes sense because light used for reading is around the visible and infrared light area (around 10E-14 to 10E-15 Hz) of the EM chart. The wave with photon energy of 2E-25 J matches the microwave, which makes sense because microwaves carry less energy than visible light, infrared, and x-rays and the energy of that photon should be less than the other activities

RRahimtoola1I
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Joined: Fri Aug 09, 2019 12:15 am

Re: HW 1.A #9

Postby RRahimtoola1I » Sat Oct 12, 2019 10:34 am

Here's an electromagnetic spectrum I found for reference. A wavelength of 340 nm would be suntanning because the sun emits UV rays. A wavelength of 1m or 1x10^9 nm would be microwaving popcorn because that wavelength falls under microwave radar. A wavelength of 2.5 nm would be x-ray because of the spectrum shown and by elimination, reading would be 600nm because it falls in the visible light area of the spectrum.
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