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An atom of helium has a de Broglie wavelength of 4.30e-12 meter. What is its velocity? The mass for 2He is 6.646e-27 kg.

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Final answer:

The velocity of the helium atom is approximately 1.46 x 10^3 m/s.

Step-by-step explanation:

The de Broglie wavelength (λ) of an object is given by the equation λ = h/mv, where h is Planck's constant (6.626 x 10^-34 Js), m is the mass of the object, and v is its velocity. To find the velocity (v) of the helium atom, we can rearrange the equation to v = h/(mλ). Plugging in the values, we have:

v = (6.626 x 10^-34 Js) / (2 x 6.646 x 10^-27 kg x 4.30 x 10^-12 m).

Simplifying the expression gives v ≈ 1.46 x 10^3 m/s.

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