Question

The position of uncertainty of an electron is 1.89 x 10 -14 m/ What is the uncertainity of velocity? thank you so much.

Answer 1

`=3.848620882*10^10 m//s`

Explanation:

For an explanation of how this question works, check out this link. Also shout out to
Truong-Son N. and his answer which helped me work this out.

To work out the uncertainty of velocity of an electron given it's uncertainty of position, we use this formula:
`Delta vec v_x=h/(m_e Delta vec x)`, where `Delta vec v_x` is the uncertainty of velocity,
`h` is Planck's constant `(6.62607004 × 10^-34 m^2 kg` `/ s`),
`m_e` is the mass of the electron (`9.10938356 × 10^-31`)
and `Delta vec x` is the uncertainty of position (in this case `1.89*10^-14m`).

Before we start I would like to note that this formula is got by transposing Heisenberg's Uncertainty Principle, which is
`Delta vec p_x Delta vec x >=h`, which transposed equals

`Delta vec p_x = h/(Delta vec x)`, then `Delta vec p_x` is swapped out for the momentum law `p=mv`, so
`Delta vec (mv_x)= h/(Delta vec x)`, divided by `m` equals

`Delta vec v_x=h/(m_e Delta vec x)`

Now we simply plug in the values into the formula, and we get
`Delta vec v_x=(6.62607004 × 10^-34 m^2 kg // s)/((9.10938356 × 10^-31kg)(1.89*10^-14m)`

`Delta vec v_x= (6.62607004 × 10^-34 (m^cancel(2))^color(red)(1) cancel(kg) // s)/(1.721673493*10^-44 cancel(kg m))`

`Delta vec v_x=3.848620882*10^10 m//s`

I hope that helped!