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

The speed of electron is 891.8
`(m)/(s)`

Step-by-step explanation:

Given :

Wavelength
`\lambda = 130 * 10^(-9)` m

According to the uncertainty principle,

`\Delta x \Delta P = (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 = (h)/(2\pi m \Delta x)`

`\Delta v = (6.626 * 10^(-34) )/(6.28 * 9.1 * 10^(-31) * 130 * 10^(-9) )` ( ∵
`h = 6.626 * 10^(-34)` )

`\Delta v =891.8`
`(m)/(s)`

Therefore, the speed of electron is 891.8
`(m)/(s)`