After an electron has been accelerated through a potential difference of 0.15 kV, what is its de Broglie wavelength? (e = 1.60 × 10^-19 C, m_electron = 9.11 × 10^-31 kg, h = 6.626 × 10^-34 J ∙ s) A) 0.10 nm B) 1.0 nm C) 1.0 mm D) 1.0 cm

Question

Quizplus · Accepted Answer

We can use the following formula to calculate the de Broglie wavelength of an electron:

λ = h / p

where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the electron, which can be calculated as follows:

p = √(2mE)

where m is the mass of the electron and E is the energy gained by the electron when it was accelerated through a potential difference of 0.15 kV. We can express E in joules using the following conversion:

1 eV = 1.60 × 10^-19 J

Therefore, the energy gained by the electron is:

E = 0.15 kV × 1 eV/1.60 × 10^-19 J/eV = 9.38 × 10^-17 J

Now we can calculate the momentum of the electron:

p = √(2mE) = √(2 × 9.11 × 10^-31 kg × 9.38 × 10^-17 J) ≈ 1.05 × 10^-24 kg m/s

Finally, we can calculate the de Broglie wavelength:

λ = h / p = 6.626 × 10^-34 J s / (1.05 × 10^-24 kg m/s) ≈ 6.31 × 10^-11 m = 0.0631 nm

Therefore, the best choice is A, 0.10 nm.

After an Electron Has Been Accelerated Through a Potential Difference