Question

What is the uncertainty in position of an electron of an atom if there is t 2.0 x 10' msec uncertainty in its velocity? Use the reduced Planck's constant and electron mass 9.19 x 103 kg.

Answer 1

18931.4

Step-by-step explanation:

Given : velocity of the electron = 2.0
`*`10

mass of the electron = 9.19
`*` 103

we know that reduced planks constant, h = 6.5821
`*`
`10^(-16)` eV s

We know from uncertainity principle,

`\Delta \textup{x}.\Delta \textup{v} = \frac{h}{\dot{m}}`

`\Delta \textup{x} = \frac{h}{\dot{m}* \Delta \textup{v}}`
`\Delta \textup{x} = (6.5821* 10^(-16))/(9.19* 103* 2.0* 10)`

`\Delta \textup{x}` = 18931.4 m

Hence, uncertainty in position of the electron is 18931.4