How many times longer is the de Broglie wavelength of a helium atom (6.647 *10^-27 kg) moving at 220 m/s than that of a baseball (0.145 kg) moving at 35 m/s?

Question

Quizplus · Accepted Answer

The de Broglie wavelength ($\lambda$) is given by $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant, $m$ is the mass, and $v$ is the velocity of the object. The ratio of the wavelengths of the helium atom to the baseball is $\frac{\lambda_{He}}{\lambda_{baseball}} = \frac{mv_{baseball}}{mv_{He}}$. Given the masses and velocities, the ratio simplifies to a constant value, indicating the helium atom's de Broglie wavelength is 3.5 times longer than that of the baseball, without needing to calculate the exact wavelengths.

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