Heisenberg Uncertainity Principle (chemistry)

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
$\Delta v= \frac {1}{2m} \sqrt {\frac {h}{\pi}}$