Question

A beam of electrons strikes a double slit, with the slit separation of 130 nm. The first-order diffraction maximum is seen at an angle of 5 degrees away from the beam's direction. What was the speed of the electrons

Answer 1

Answer:

The speed of electron is 891.8 $\frac{m}{s}$

Explanation:

Given :

Wavelength $\lambda = 130 \times 10^{-9}$ m

According to the uncertainty principle,

    $\Delta x \Delta P = \frac{h}{2\pi }$

Where $\Delta x =$ difference in length means wavelength, $\Delta P =$ difference in momentum,

We know that $\Delta P = m\Delta v$

Put value in main equation,

 $\Delta v = \frac{h}{2\pi m \Delta x}$

 $\Delta v = \frac{6.626 \times 10^{-34} }{6.28 \times 9.1 \times 10^{-31} \times 130 \times 10^{-9} }$                   ( ∵ $h = 6.626 \times 10^{-34}$ )

 $\Delta v =891.8$ $\frac{m}{s}$

Therefore, the speed of electron is 891.8 $\frac{m}{s}$