Respuesta :
Answer:
Minimum uncertainty in position is [tex]\Delta x= 1157808.48\ m[/tex]
Explanation:
It is given that,
Uncertainty in the position of an electron, [tex]\Delta x=0.1\ nm=0.1\times 10^{-9}\ m[/tex]
According to uncertainty principle,
[tex]\Delta x.\Delta p\geq \dfrac{h}{4\pi}[/tex]
[tex]\Delta x.m\Delta v\geq \dfrac{h}{4\pi}[/tex]
[tex]\Delta v\geq \dfrac{h}{4\pi \times \Delta x\times m}[/tex]
[tex]\Delta v\geq \dfrac{6.62\times 10^{-34}\ J-s}{4\pi \times 0.1\times 10^{-9}\ m\times 9.1\times 10^{-31}\ kg}[/tex]
[tex]\Delta v\geq 578904.24\ m/s[/tex]
Let [tex]\Delta x[/tex] is the uncertainty in position after 2 seconds such that,
[tex]\Delta x=\Delta v\times t[/tex]
[tex]\Delta x=578904.24\ m/s\times 2\ s[/tex]
[tex]\Delta x= 1157808.48\ m[/tex]
or
[tex]\Delta x= 1.15\times 10^6\ m[/tex]
Hence, this is the required solution.
