Answer:
The uncertainty in the position of the electron is [tex]5.79x10^{-9}m[/tex]
Explanation:
The Heisenberg uncertainty principle is defined as:
[tex]\Lambda p\Lambda x[/tex] ≥ [tex]\frac{h}{4 \pi}[/tex] (1)
Where [tex]\Lambda p[/tex] is the uncertainty in momentum, [tex]\Lambda x[/tex] is the uncertainty in position and h is the Planck's constant.
The momentum is defined as:
[tex]p =mv[/tex] (2)
Therefore, equation 2 can be replaced in equation 1
[tex]\Lambda (mv) \Lambda x[/tex] ≥ [tex]\frac{h}{4 \pi}[/tex]
Since, the mass of the electron is constant, v will be the one with an associated uncertainty.
[tex]m \Lambda v \Lambda x[/tex] ≥ [tex]\frac{h}{4 \pi}[/tex] (3)
Then, [tex]\Lambda x[/tex] can be isolated from equation 3
[tex]\Lambda x[/tex] ≥ [tex]\frac{h}{m \Lambda v 4 \pi}[/tex] (4)
[tex]\Lambda x = \frac{6.626x10^{-34}J.s}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) 4 \pi}[/tex]
But [tex]1J = Kg.m^{2}/s^{2}[/tex]
[tex]\Lambda x = \frac{(6.624x10^{-34} Kg.m^{2}/s^{2}.s)}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) (4 \pi)}[/tex]
[tex]\Lambda x = 5.79x10^{-9}m[/tex]
Hence, the uncertainty in the position of the electron is [tex]5.79x10^{-9}m[/tex]
