Heisenberg Uncertainty Principle
According to this, it is impossible to determine simultaneously the exact position & the exact momentum (or velocity) of an electron
Mathematically:
- Δx is uncertainty in position Δ v is in velocity.
- If the position of the electron is known with high degree of accuracy ($\Delta x$ is small), then the velocity of the electron
will be uncertain [$\Delta (v_x)$ is large]. On the other hand, if the velocity of the electron is known precisely ($\Delta (v_x )$ is small), then the position of
the electron will be uncertain ($\Delta x$ will be large).
- So, if we carry out some physical measurements on the electron's position or velocity, the outcome will always depict a fuzzy or blur picture
- The effect of this principle is only significant for microscopic objects and it is negligibile for macroscopic objects
Question 1
If the position of electron is measured with accuracy of 0.002 nm. Calculate uncertainty in momentum of e
Solution
$\Delta x. \Delta p = \frac {h}{4 \pi}$
$0.002 \times 10^(-9 ).\Delta p= \frac {6.6 \times 10^{-34}}{4 \times 3.14}$
$ \Delta p= \frac {6.6 \times 10^{-34}}{4 \times 3.14 \times 0.002 \times 10^{-9} }$
= 2.637 X 10-23 kg m/s
Question 2
A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 A0. What is the uncertainty involved in the measurement of its velocity?
Solution
$\Delta x. \Delta p = \frac {h}{4 \pi}$
or
$\Delta x. m \Delta v = \frac {h}{4 \pi}$
$\Delta v= \frac {h}{4 m \pi \Delta x}$
Now $\Delta x= .1 A^0 = .1 \times 10^{-10}$ m
Substituting all the values
$\Delta v = \frac {6.626 \times 10^{-34}}{4 \times 3.14 \times .1 \times 10^{-10} \times 9.11 \times 10^{-31}}$
$=5.79 \times 10^6$ m/s
Question 3
A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position. ?
Solution
The uncertainty in the speed is 2%, i.e.
$\Delta v= 45 \times \frac {2}{100} = .9$ m/s
Now
$\Delta x= \frac {h}{4 m \pi \Delta v}$
Substituting all the values, we get
$\Delta x=1.46 \times 10^{-33}$ m
Question 4
If uncertainty in position of electron is zero, the uncertainty in its momentum would be
(i)zero
(ii) $ > \frac {h}{4 \pi}$
(iii) $ < \frac {h}{2 \pi} $
(iv)infinity
Solution
(iv)
Question 5
if uncertainty in position and momentum are equal then uncertainty in velocity is
(i)$ \sqrt {\frac {h}{2 \pi}}$
(ii) $ \frac {1}{2m} \sqrt {\frac {h}{\pi}}$
(iii)$ \sqrt {\frac {h}{ \pi}}$
(iv)None
Solution
$\Delta x. \Delta p = \frac {h}{4 \pi}$
The uncertainty in position and momentum are equal.let it be y,then
$y^2 = \frac {h}{4 \pi}$
$y= \sqrt {\frac {h}{4 \pi}}$
Now
$m \Delta v = \sqrt {\frac {h}{4 \pi}}$
or
$\Delta v= \frac {1}{2m} \sqrt {\frac {h}{\pi}}$
Significance of Uncertainty Principle:-
-
It rules out definite paths or orbits of electrons and other similar particles.
-
This principle is significant only for microscope object like electrons and negligible for macroscopic objects
Reason for failure of Bohr model:-
-
The wave character is not considered in Bohr model.
-
According to Bohr model, an orbit is clearly defined path & this path can completely be defined only if both position & velocity of electron are known as same lime. This is not possible according to Heisenberg uncertainty principle.
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