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 A^{0}. 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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