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QuestionCalculate the de Broglie wavelength of an electron moving at 2.35×106 m/s2.35 \times 10^{6} \mathrm{~m} / \mathrm{s}.

Studdy Solution

STEP 1

Assumptions1. The speed of the electron is .35×106 m/s.35 \times10^{6} \, \mathrm{m/s} . We will use the de Broglie formula to calculate the wavelength3. The mass of the electron is known and constant 9.11×10−31 kg9.11 \times10^{-31} \, \mathrm{kg}
4. Planck's constant is known and constant 6.626×10−34 mkg/s6.626 \times10^{-34} \, \mathrm{m^ kg / s}

STEP 2

The de Broglie wavelength of a particle is given by the formulaλ=hmv\lambda = \frac{h}{mv}where λ\lambda is the wavelength, hh is Planck's constant, mm is the mass of the particle, and vv is the velocity of the particle.

STEP 3

Now, we plug in the given values for Planck's constant, the mass of the electron, and the velocity of the electron into the formula.
λ=6.626×10−34 m2kg/s9.11×10−31 kg×2.35×106 m/s\lambda = \frac{6.626 \times10^{-34} \, \mathrm{m^2 kg / s}}{9.11 \times10^{-31} \, \mathrm{kg} \times2.35 \times10^{6} \, \mathrm{m/s}}

STEP 4

Perform the multiplication in the denominator.
λ=6.626×10−34 m2kg/s2.14×10−24 kg⋅m/s\lambda = \frac{6.626 \times10^{-34} \, \mathrm{m^2 kg / s}}{2.14 \times10^{-24} \, \mathrm{kg \cdot m/s}}

STEP 5

Finally, divide the numerator by the denominator to find the wavelength.
λ=.626×10−34 m2kg/s2.14×10−24 kg⋅m/s=3.10×10−10 m\lambda = \frac{.626 \times10^{-34} \, \mathrm{m^2 kg / s}}{2.14 \times10^{-24} \, \mathrm{kg \cdot m/s}} =3.10 \times10^{-10} \, \mathrm{m}The de Broglie wavelength of an electron traveling with a speed of 2.35×10 m/s2.35 \times10^{} \, \mathrm{m/s} is 3.10×10−10 m3.10 \times10^{-10} \, \mathrm{m}.

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