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Rutherford used a beam of alpha particles for his scattering experiment what is the de Broglie wavelength of an alpha particle moving at a velocity of 2.15×10^7 m per second

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Final answer:

The de Broglie wavelength of an alpha particle moving at a velocity of 2.15 × 10^7 m/s is 4.93 × 10^-15 m.

Step-by-step explanation:

The de Broglie wavelength of a particle can be calculated using the equation:

λ = h/mv

Where λ is the de Broglie wavelength, h is the Planck's constant (6.626 × 10^-34 Js), m is the mass of the particle, and v is the velocity of the particle.

In the case of an alpha particle with a velocity of 2.15 × 10^7 m/s, the mass of the alpha particle is 6.64 × 10^-27 kg. Plugging these values into the equation, we get:

λ = (6.626 × 10^-34 Js) / (6.64 × 10^-27 kg) × (2.15 × 10^7 m/s) = 4.93 × 10^-15 m

Therefore, the de Broglie wavelength of an alpha particle moving at a velocity of 2.15 × 10^7 m/s is 4.93 × 10^-15 m.

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