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Topic: Last one dealing with de Broglie  (Read 5182 times)

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Offline achibaby1974

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Last one dealing with de Broglie
« on: April 05, 2008, 07:00:10 PM »
The final one:

An alpha particle (mass = 6.6 x 10 ^ -24 g) emitted by radium travels at 3.4 x 10 ^ 7   +/- (plus or minus) 0.1 x 10 ^ 7 mi/h. What is its de Broglie wavelength in meters?

De Broglie: wavelength = (Planck’s constant) / (mass x velocity)

What’s the deal with the +/-? What does that mean? Is there a range for the answer? The only thing I can think of is to solve both ways and see which one yields a (-) answer because it says the particle is emitted meaning it loses energy, meaning exothermic. But we're talking about wavelength so I'm confused. And what about the radium? Is there any significance to it?           

Thank you for any help at all. I have a test on Monday and I can't figure these 3 out.

Offline macman104

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Re: Last one dealing with de Broglie
« Reply #1 on: April 05, 2008, 07:06:20 PM »
the +/- is just an error bound.  So you would, I imagine solve for both ends of the possibility, 3.3x10^7 and 3.5x10^7.  The fact that it is Radium is just filler information, it could be any element.  This is a pretty straightforward "plug and chug" problem.

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