What is the predicted de Broglie wavelength of a bullet traveling at 1060 m/s?

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SUMMARY

The predicted de Broglie wavelength of a bullet with a mass of 39 grams traveling at a speed of 1060 m/s is calculated using the de Broglie equation: wavelength = h / p. The momentum (p) is determined by the formula p = mv, resulting in a value of 4.134 x 10^-2 kg*m/s. The correct de Broglie wavelength is approximately 1.604 x 10^-34 meters, confirming that the bullet's wavelength is extremely small and not observable as a matter wave.

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Homework Statement


A bullet of mass 39 g travels at 1060 m/s. Although the bullet is clearly too large to be treated as a matter wave, determine what Eq. 38-13 predicts for the de Broglie wavelength of the bullet at that speed?


Homework Equations


de Broglie equation: wavelength = h / p where h is Planck constant and p is the particle's momentum magnitude.


The Attempt at a Solution


p=sqrt(2*m*K) where K=1/2mv^2. I solved K first and got 21,910,200. Then I solved p and got 41,340. Then I solved for wavelength using the de Broglie eq'n and got 1.604 x 10^-38, but this is not the correct answer. Where did I go wrong?? Thanks for your help!
 
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Huh, I'd start by solving for the momentum (3.9 * 10^-2 kg)(1.060 * 10^3 m/s) = 4.134 *10 kg*m/s, then I would recall that p = h/wavelength and solve for the wavelength. It should be a really big number (of the order 10^34 in the mks system).

Good luck.
 
Yeah, hence it becomes 3.9 times 10^-2 kilograms.
 

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