Question 6
1 answer:
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(A)
The De Broglie wavelength of an electron is given by
(1)
where
h is the Planck constant
p is the momentum of the electron
The electron in this problem has a speed of
and its mass is
So, its momentum is
And substituting into (1), we find its De Broglie wavelength
(B)
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
And so, the De Broglie wavelength of the ball is given by
(C)
The location of the first intensity minima is given by
where in this case we have
L = 1.091 is the distance between the detector and the slit
is the width of the slit
Solving the formula for , we find the wavelength of the electrons in the beam:
(D)
The momentum of one of these electrons can be found by re-arranging the formula of the De Broglie wavelength:
where here we have
is the wavelength
Substituting into the formula, we find