Answer:
[tex]9.9\cdot 10^{-9}m[/tex]
Explanation:
In a single-slit diffraction pattern, the location of the first minimum is given by
[tex]y=\frac{L\lambda }{a}[/tex]
where
L is the distance between the slit and the screen
[tex]\lambda[/tex] is the wavelength
a is the width of the slit
In this problem, we have
[tex]y=0.527 cm = 5.27\cdot 10^{-3} m[/tex] (location of first minimum)
L = 1.068 m (distance of the screen)
[tex]a=2.00\mu m=2.00\cdot 10^{-6}m[/tex] (width of the slit)
Solving the equation for [tex]\lambda[/tex], we find the De Broglie wavelength of the electron:
[tex]\lambda = \frac{ya}{L}=\frac{(5.27\cdot 10^{-3} m)(2.00\cdot 10^{-6} m)}{1.068 m}=9.9\cdot 10^{-9}m[/tex]