Approximately, what is the de Broglie wavelength of an electron that has been accelerated through a potential difference of 225 V? The mass of an electron is 9.11 × 10^-³¹ kg. A)0.14 nm B)0.082 nm C)0.043 nm D)0.0092 nm E)0.22 nm

Question

Quizplus · Accepted Answer

The de Broglie wavelength of a particle is given by the equation:

λ = h / p

where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle.

The momentum of an electron that has been accelerated through a potential difference of V volts is given by:

p = √(2m_eV)

where m_e is the mass of the electron.

Substituting the given values, we get:

p = √(2 * 9.11 × 10^-31 kg * 225 V) = 1.77 × 10^-24 kg m/s

Substituting this value into the de Broglie wavelength equation, we get:

λ = h / p = 6.63 × 10^-34 J s / 1.77 × 10^-24 kg m/s = 0.082 nm

Therefore, the best choice is B, with a de Broglie wavelength of 0.082 nm.

Approximately, What Is the De Broglie Wavelength of an Electron