Question

An electron has a speed of 500 m$s^{-1}$ with an uncertainty of 0.02%. What is the uncertainty in locating its position?

Answer 1

To determine: The uncertainty in locating the position of an electron

Given Data : Speed of Electron = 500 m/s, Uncertainty in Speed = 0.02%

Formulas to be used:

1) Heisenberg uncertainty principle:

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

Where $\Delta x$ = uncertainty in position

$\Delta p$ = uncertainty in momentum

Planck's constant, h\quad $=\quad 6.63\quad \times \quad { 10 }^{ -34 }\quad kg\sfrac { { m }^{ 2 } }{ s }$

2) For calculating \Delta p

$\Delta p\quad =\quad m\Delta v$

Where m is the mass of electron

$\Delta v$ = uncertainty in speed

Calculation:

Step 1: Determine $\Delta v$

$\Delta v\quad =\quad \frac { 0.02 }{ 100 } \quad \times \quad 500\quad =\quad 0.1\quad \sfrac { m }{ s }$

Step 2: Calculate $\Delta p$

$\Delta p\quad =\quad m\Delta v$

We know that, $m\quad =\quad 9.1\quad \times \quad { 10 }^{ -31 }\quad kg$

Substituting the values of m and   (obtained from step 1), we get

$\Delta p\quad =\quad \left( 9.1\quad \times \quad { 10 }^{ -31 } \right) \quad \times \quad \left( 0.1 \right) \quad =\quad 9.1\quad \times \quad { 10 }^{ -32 }\quad kg\sfrac { m }{ s }$

Step 3: Apply Heisenberg uncertainty principle

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

$\Delta x\quad \ge \quad \frac { h }{ 4\pi } \quad \times \quad \frac { 1 }{ \Delta p }$

$\frac { h }{ 4\pi } \quad \times \quad \frac { 1 }{ \Delta p } \quad =\quad \frac { 6.63\quad \times \quad { 10 }^{ -34 } }{ 4\quad \times \quad 3.14 } \quad \times \quad \frac { 1 }{ 9.1\quad \times \quad { 10 }^{ -32 } } \quad =\quad 5.8\quad \times \quad { 10 }^{ -4 }\quad m$

Hence, $\Delta x\quad \ge \quad 5.8\quad \times \quad { 10 }^{ -4 }\quad m$