Question

An electron has a speed of 1.05×10^4ms-1 within accuracy of 0.01%. calculate the uncertainty in the position of an electron

Answer 1

Answer:

Using Heisenberg's Uncertainty Principle,

   Δx.Δv=h/4πm

         Δx=h/4πmΔv

              =6.626x10^-34/4x3.14x9.1x10^-31x1.05x10^4

              ≈4.96 x 10^-60

HOPE IT HELPS YOU

         

Answer 2

Answer:

The uncertainty in the position of an electron is $5.59\times 10^-^7m$.

Explanation:

We will use Heisenberg's uncertainty principle to solve this question. It is given as,

$\Delta x.m\Delta v=\frac{h}{4\pi}$           (1)

Where,

Δx=uncertanity in the position of the moving particle

m=mass of the moving particle

Δv=uncertanity in velocity of the particle

h=planks constant=6.6×10⁻³⁴

From the question we have,

Speed(v)=1.05×10⁴m/s

Speed is accurate upto=0.01%

Uncertainty in speed is=(1-0.01)=0.99%

Mass of electron is=9.1×10⁻³¹kg

So, the uncertain speed is,

$\Delta v=0.99\times 10^-^2\times 1.05\times 10^4=1.0395\times 10^2m/s$

By substituting the required values in equation (1) we get;

$\Delta x\times 9.1\times 10^-^3^1\times 1.0395\times 10^2=\frac{6.6\times 10^-^3^4}{4\pi}$

$\Delta x=\frac{6.6\times 10^-^3^4}{4\pi\times 9.1\times 10^-^3^1\times 1.0395\times 10^2}$             (π=3.14)

$\Delta x=5.59\times 10^-^7m$

Hence, the uncertainty in the position of an electron is $5.59\times 10^-^7m$.