# Chapter 6 - Electronic Structure of Atoms - Exercises: 6.47b

The wavelength of the bullet is $\lambda=2.65\times10^{-34}m$.

#### Work Step by Step

*The de Broglie relationship: $$\lambda=\frac{h}{mv}$$ $\lambda$: wavelength of object $h$: Planck's constant ($h=6.626\times10^{-34} J.s$) $m$: mass of object $v$: velocity of object 1) Find the known variables - Mass of the bullet: $m=10.0g=10^{-2}kg$ - Velocity of the bullet: $v=250m/s$ 2) Calculate the wavelength of the bullet $$\lambda=\frac{h}{mv}=\frac{6.626\times10^{-34}}{10^{-2}\times250}\approx2.65\times10^{-34}m$$

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