/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 47 The electron interference patter... [FREE SOLUTION] | 91Ó°ÊÓ

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The electron interference pattern of Figure 38.12 was made by shooting electrons with 50keVof kinetic energy through two slits spaced role="math" localid="1650737433408" 1.0μ³¾apart. The fringes were recorded on a detector 1.0mbehind the slits.

a. What was the speed of the electrons? (The speed is large enough to justify using relativity, but for simplicity do this as a nonrelativistic calculation.)

b. Figure 38.12 is greatly magnified. What was the actual spacing on the detector between adjacent bright fringes?

Short Answer

Expert verified

(a) The speed of the electron isv=1.33×108ms

(b) The actual distance between consecutive light fringes on the detector isΔ²â=5.47μ³¾

Step by step solution

01

Given information

The kinetic energy of electrons is 50 KeV and a detector captured the fringes is 1.0 m

02

Part (a) Step 1: Given information

The electron starts at rest (or close to it), thus the kinetic energy obtained is12mv2

03

Part (a) Step 2: Finding the electron's speed using kinetic energy 

We can begin by calculating the electron's speed using kinetic energy,

Ek=mv22v=2Ekmexpressvv=2⋅50×103×1.6×10−199.11×10−31v=1.33⋅108ms

04

Part (b) Step 1: Given information

On a detector 1.0 m behind the slits, the fringes were recorded

05

Part (b) Step 2: Determine the double slit experiment, fringe spacing 

In a double slit experiment, we can now utilise an expression for fringe spacing.

Δ²â=³¢Î»dλ=hmv(de Broglie wavelength)Δ²â=Lhmvd(substituteλ)Δ²â=1â‹…6.63×10−349.11×10−31×1.33×108×1×10−6(substitute)Δ²â=5.47μ³¾

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