An electron (m_e = 9.11 $\times$ 10^-31 kg)traveling with a velocity of 1.1 $\times$ 10⁶ m/s has a de Broglie wavelength of approximately A)0.66 nm B)0.78 nm C)0.89 nm D)0.97 nm E)1.2 nm

Question

Quizplus · Accepted Answer

The de Broglie wavelength ($\lambda$) of a particle is given by the formula $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{m}^2 \text{kg} / \text{s}$), $m$ is the mass of the particle, and $v$ is the velocity of the particle. Substituting the given values ($m = 9.11 \times 10^{-31} \, \text{kg}$ and $v = 1.1 \times 10^6 \, \text{m/s}$), we can calculate the de Broglie wavelength. The calculation yields a wavelength of approximately $0.66 \, \text{nm}$, making option A the correct answer.

An Electron (Me = 9 ×\times× 10-31 Kg)traveling with a Velocity of 1

An Electron (M_e = 9 $\times$ 10^-31 Kg)traveling with a Velocity of 1