Question
Uncertainty in the position of an electron $$\left( {mass = 9.1 \times {{10}^{ - 31}}kg} \right)$$ moving with a velocity $$300\,m{s^{ - 1}},$$ accurate upto $$0.001\% $$ will be $$\left( {h = 6.63 \times {{10}^{ - 34}}Js} \right)$$
A.
$$1.92 \times {10^{ - 2}}m$$
B.
$$3.84 \times {10^{ - 2}}m$$
C.
$$19.2 \times {10^{ - 2}}m$$
D.
$$5.76 \times {10^{ - 2}}m$$
Answer :
$$1.92 \times {10^{ - 2}}m$$
Solution :
$${\text{Given}}\,\,m = 9.1 \times {10^{ - 31kg}},$$ $$h = 6.6 \times {10^{ - 34}}Js$$
$$\eqalign{
& \Delta v = \frac{{300 \times .001}}{{100}} \cr
& \,\,\,\,\,\,\,\,\, = 0.003m{s^{ - 1}} \cr} $$
From Heisenberg's uncertainity principle
$$\eqalign{
& \Delta x = \frac{{6.62 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 0.003 \times 9.1 \times {{10}^{ - 31}}}} \cr
& \,\,\,\,\,\,\,\,\, = 1.92 \times {10^{ - 2}}m \cr} $$