How many times longer is the de Broglie wavelength of a helium atom (6.647 $\times$ 10^-27kg)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 is given by the equation λ = h/p, where h is Planck's constant and p is the momentum of the particle. 
For the helium atom: 
momentum = mass x velocity = 6.647 x 10^-27 kg x 220 m/s 
λ = h/p = (6.626 x 10^-34 J s) / (6.647 x 10^-27 kg x 220 m/s) 
λ = 3.5 x 10^-10 m

For the baseball: 
momentum = mass x velocity = 0.145 kg x 35 m/s 
λ = h/p = (6.626 x 10^-34 J s) / (0.145 kg x 35 m/s) 
λ = 5 x 10^-35 m

The ratio of the two wavelengths is: 
(3.5 x 10^-10 m) / (5 x 10^-35 m) = 7 x 10^24

Therefore, the de Broglie wavelength of the helium atom is 7 x 10^24 times longer than that of the baseball. The closest option is A) 3.5 x 10^10, which is 3.5 x 10^24 in scientific notation.

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