Question

State Heisenberg’s uncertainty principle. An electron has a velocity of 50 m$s^{-1}$ accurate upto 99.99%. Calculate the uncertainty in locating its position. (Mass of electron = $9.1 \times 10^{-31}$ kg, h = $6.6 \times 10^{-34}$ Js)

Answer 1

"Heisenberg Uncertainty Principle states that is fundamentally impossible to know precisely both the velocity and position of a particle at the same time.

$\Delta x.\Delta p\quad \ge \quad \frac { h }{ 4\pi  }$

$\Delta x\quad =\quad Uncertainty\quad of\quad position$

$\Delta p\quad =\quad Uncertainty\quad of\quad momentum$

From the given,

$\Delta v\quad =\quad \frac { 50\quad \times \quad 99.99 }{ 100 } \quad =\quad 49.995\quad m/s\quad$

$m\quad =\quad 9.1\times { 10 }^{ -31 }\quad kg\quad$

$h\quad =\quad 6.6\times { 10 }^{ -34 }\quad Js$

$\Delta \chi \quad =\quad \frac { h }{ 4\pi m\Delta v }$

$=\quad \frac { 6.626\quad \times \quad { 10 }^{ -34 } }{ 4\quad \times \quad 3.14\quad \times \quad 10\quad \times \quad { 10 }^{ -3 }\quad \times \quad 49.99 }$

$\Delta \chi \quad =\quad 1.46\quad \times \quad { 10 }^{ -33 }\quad m$"