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What is the de Broglie wavelength of an electron traveling at 1.02×105m/s?

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Answer:

1.82 x 10^-12 m.

Step-by-step explanation:

The de Broglie wavelength of an electron traveling at 1.02 x 10^5 m/s is 1.82 x 10^-12 m.

The de Broglie wavelength of a particle is a measure of the particle's wave-like behavior, and it is given by the following equation:

λ = h / mv

where λ is the de Broglie wavelength, h is the Planck constant, m is the mass of the particle, and v is the velocity of the particle.

In the case of an electron traveling at 1.02 x 10^5 m/s, the de Broglie wavelength is given by the following calculation:

λ = 6.62 x 10^-34 J * s / (9.11 x 10^-31 kg * 1.02 x 10^5 m/s)

This simplifies to:

λ = 1.82 x 10^-12 m

Therefore, the de Broglie wavelength of an electron traveling at 1.02 x 10^5 m/s is 1.82 x 10^-12 m.

User Thibaut Balabonski
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