Question

An electron has a speed $3\times {{10}^{2}}m{{s}^{-1}}$ with uncertainty 0.07 %. What is the uncertainty in locating its position.

Answer 1

Hint:. We have a relation,$\Delta v = (3\times {{10}^{2}})\times 0.07%$
The value of Planck's constant, $h = 6.626\times {{10}^{-34}}Js$

Complete step by step answer:
So the question is about the Heisenberg’s uncertainty principle, where the value for the velocity of electrons is given as $3\times {{10}^{2}}m{{s}^{-1}}$ and with and uncertainty 0.07 %. And we have to find the uncertainty in position of the electron from the given data.
- So let’s write the equation for Heisenberg’s uncertainty principle. The equation is as follows $\Delta x\times m\times \Delta v=\dfrac{h}{4\pi }$
Where $\Delta x$ = uncertainty in position, $\Delta v$ = uncertainty in velocity
m = mass of the electron
h = Plank’s constant
- So here for us, velocity is given and uncertainty value is also given from this we have to find the uncertainty in velocity i.e. $\Delta v$, to find the uncertainty in position ($\Delta x$).
To find the value of $\Delta v$, the equation is,
$\text{ }\!\!\Delta\!\!\text{ v = velocity}\,\text{of}\,\text{electron }\!\!\times\!\!\text{ }\dfrac{\text{uncertainty}\,\text{value}}{\text{100}}$
- Here uncertainty value is given in percentage, so it has to be divided with 100.
$\text{ }\!\!\Delta\!\!\text{ v = 300 }\!\!\times\!\!\text{ }\dfrac{0.07}{\text{100}}$
$\text{ }\!\!\Delta\!\!\text{ v = 0}\text{.21m/s}$
So we got the value of$\Delta v$, so now let’s substitute all the values in the uncertainty principle.$\text{ }\!\!\Delta\!\!\text{ v=0}\text{.21m/s}$
m = $9.1\times {{10}^{-31}}$kg
$h = 6.626\times {{10}^{-34}}Js$
$\pi = 3.14$
- Substitute all these values in the uncertainty principle equation,
$\Delta x\times m\times \Delta v =\dfrac{h}{4\pi }$
$\Delta x\times 9.1\times {{10}^{-31}}\times 0.21=\dfrac{6.626\times {{10}^{-34}}}{4\times 3.14}$
Now, alter this equation and solve for uncertainty of position ($\Delta x$).
$\Delta x = \dfrac{6.626\times {{10}^{-34}}}{4\times 3.14}\times \dfrac{1}{9.1\times {{10}^{-31}}\times 0.21}$
$\Delta x = 2.7605\times {{10}^{-4}}m$
So the value of uncertainty in locating the position of the electron is $2.7605\times {{10}^{-4}}m$.

Note: If in the question mass and velocity is given, we know the mass of the electron, but if velocity is not given and we are only provided with the value of momentum of the electron, then the equation we use is:
$\Delta x\times \Delta p=\dfrac{h}{4\pi }$
Where p is the momentum of the electron.