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QuestionFind the de Broglie wavelength of an electron moving at 2.35×106 m/s2.35 \times 10^{6} \mathrm{~m/s}. Choices: 5.14×10−37 m5.14 \times 10^{-37} \mathrm{~m}, 2.74×10−10 m2.74 \times 10^{-10} \mathrm{~m}, 3.10×10−10 m3.10 \times 10^{-10} \mathrm{~m}, 3.23×109 m3.23 \times 10^{9} \mathrm{~m}.

Studdy Solution

STEP 1

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

STEP 2

The de Broglie wavelength of a particle is given by the equationλ=hmv\lambda = \frac{h}{mv}whereλ\lambda is the de Broglie wavelength, hh is the Planck's constant, mm is the mass of the particle, andvv is the velocity of the particle.

STEP 3

Now, plug in the given values for the Planck's constant, the mass of the electron, and the velocity of the electron to calculate the de Broglie wavelength.
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 kgm/s\lambda = \frac{6.626 \times10^{-34} \mathrm{~m^2 kg / s}}{2.14 \times10^{-24} \mathrm{~kg m/s}}

STEP 5

Divide the numerator by the denominator to calculate the de Broglie wavelength.
λ=.626×10−34 m2kg/s2.14×10−24 kgm/s=3.10×10−10 m\lambda = \frac{.626 \times10^{-34} \mathrm{~m^2 kg / s}}{2.14 \times10^{-24} \mathrm{~kg 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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