Electrons with a speed of 2.1 × 10⁶ m/s are directed towards a 1.0-μm wide slit.An electron detector is placed 1.0 m behind the slit.How wide is the central maximum of the electron diffraction pattern on the detector? (h = 6.626 × 10^-34 J ∙ s,m_electron = 9.11 × 10^-31 kg) A)690 μm B)350 μm C)1,000 μm D)1,400 μm

Question

Quizplus · Accepted Answer

The width of the central maximum ($Δy$) in a diffraction pattern can be calculated using the formula $Δy = 2Lλ/d$, where $L$ is the distance from the slit to the screen (1.0 m), $λ$ is the de Broglie wavelength of the electrons, and $d$ is the width of the slit (1.0 μm). The de Broglie wavelength $λ = h/p$, where $h$ is Planck's constant and $p$ is the momentum of the electron. The momentum $p = mv$, with $m$ being the mass of the electron and $v$ its velocity. Substituting the given values, $λ = (6.626 × 10^{-34}) / (9.11 × 10^{-31} × 2.1 × 10^6)$, and then calculating $Δy$, results in a width of approximately 690 μm for the central maximum.

Electrons with a Speed of 2