Question

Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed is low enough that the classical momentum formula p=mv is still valid.) Recall that the mass of an electron is m_{\rm e} = 9.11\times 10^{-31}\; {\rm kg}, and Planck's constant is h = 6.626 \times 10^{-34}\; {\rm J \cdot s}.
Express your answer in meters to three significant figures.
lambda =7.270×10−10 \rm m

Part B

Find the de Broglie wavelength lambda of a baseball pitched at a speed of 40.0 m/s. Assume that the mass of the baseball is 0.143 \;{\rm kg}.
Express your answer in meters to three significant figures

lambda =1.16×10−34 \rm m


As a comparison, an atomic nucleus has a diameter of around 10^{-14}\;{\rm m}. Clearly, the wavelength of a moving baseball is too small for you to hope to see diffraction or interference effects during a baseball game.

Part C

Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00 \;\mu{\rm m}. The electrons then head toward an array of detectors a distance 1.091 m away. These detectors indicate a diffraction pattern, with a broad maximum of electron intensity (i.e., the number of electrons received in a certain area over a certain period of time) with minima of electron intensity on either side, spaced 0.492 cm from the center of the pattern. What is the wavelength lambda of one of the electrons in this beam? Recall that the location of the first intensity minima in a single slit diffraction pattern for light is y=L \lambda /a, where L is the distance to the screen (detector) and a is the width of the slit. The derivation of this formula was based entirely upon the wave nature of light, so by de Broglie's hypothesis it will also apply to the case of electron waves.

Express your answer in meters to three significant figures.

lambda =9.02×10−9 \rm m

Part D

What is the momentum p of one of these electrons?
Express your answer in kilogram-meters per second to three significant figures.

Answer 1

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the electron

The electron in this problem has a speed of

$v=1.00\cdot 10^6 m/s$

and its mass is

$m=9.11\cdot 10^{-31} kg$

So, its momentum is

$p=mv=(9.11\cdot 10^{-31} kg)(1.00\cdot 10^6 m/s)=9.11\cdot 10^{-25}kg m/s$

And substituting into (1), we find its De Broglie wavelength

$\lambda=\frac{6.63\cdot 10^{-34}Js}{9.11\cdot 10^{-25} kg m/s}=7.28\cdot 10^{-10} m$

(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

$\lambda=\frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{5.72 kg m/s}=1.16\cdot 10^{-34}m$

(C) $9.02\cdot 10^{-9}m$

The location of the first intensity minima is given by

$y=\frac{L\lambda}{a}$

where in this case we have

$y=0.492 cm = 4.92\cdot 10^{-3} m$

L = 1.091 is the distance between the detector and the slit

$a=2.00\mu m=2.00\cdot 10^{-6}m$ is the width of the slit

Solving the formula for $\lambda$, we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:

$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$