Find the wave length of a 2.0 g light ball moving with a velocity: (a) 1.0 mm per century (b) 1.0 m/s Given: h=6.63*10^-34 Js and 1yr=3.15*10^7s
I think you might have to use a form of binomial expansion for numbers that small. Is that what this is practice for? I assume you meant matter wave. I went to Wikipedia ( http://en.wikipedia.org/wiki/Matter_wave ) and found this:\[\lambda=\frac{h}{\gamma\ m_0\ v}\]
I think it was binomial expansion, but I'm not quite 100% sure.
γ =? and mov=?
Is this for special relativity?
this problem is of relativity
If we're trying to find the wavelength of a matter wave, I think that equation from Wikipedia is the answer. And is the ball's rest mass \(2.0\ [g]\)? Or is that the relativistic mass?
the answer is (a) 1.05*10^-18m and (b) 3.3*10^-31m
\[1.14345\times 10^{-18}\ [m]\div\gamma\]\[3.315\times 1^{-31}\ [m]\div\gamma\](according to my Windows calculator) Where \(\gamma\approx 1\) due to low speed compared to the speed of light. So, it appears the equation from Wikipedia is correct. So, are you currently trying to understand this problem?
yes
Okay. Well, I just used that equation. Do you know about the Lorentz factor, represented by \(\gamma\)?
i don't know, is it frequency?
Actually, nope! It's a special value, though. I don't know where it comes from, but it's very important in many equations in special relativity. The variable is pronounced "gamma," and \(\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\).
That's just a good thing to remember. It's important. \(m_0\) is "rest mass." As a mass gets closer to the speed of light, mass increases. It's least mass is the mass it has when at rest. That means there is only one rest mass, and it is pretty important. All the mass you measure here on Earth is pretty much the rest mass. \(v\) is the relative velocity, and that's what you're given in (a) and (b). \(h\) is another very important number. But it doesn't have a formula that you need to use, because it is constant. \(h\) is given to you in the problem, for your convenience.
h= 6.63*10^-34 J.s
Right.
And you need \(v\) to be in meters per second for the units to work out right.
yes
Right. So do you understand how to get the answer, now?
yes thank you so much for your help
You're welcome! I'm glad I could help. Good luck with everything you do!
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