Question
In an atom, an electron is moving with a speed of $$600\,m/s$$ with an accuracy of $$0.005\% .$$ Certainity with which the position of the electron can be located is ( $$h = 6.6 \times {10^{ - 34}}kg\,{m^2}{s^{ - 1}},$$ mass of electron, $${e_m} = 9.1 \times {10^{ - 31}}kg$$ ) :
A.
$$5.10 \times {10^{ - 3}}m$$
B.
$$1.92 \times {10^{ - 3}}m$$
C.
$$3.84 \times {10^{ - 3}}m$$
D.
$$1.52 \times {10^{ - 4}}m$$
Answer :
$$1.92 \times {10^{ - 3}}m$$
Solution :
$$\eqalign{
& {\text{According to Heisenberg uncertainty principle}}{\text{.}} \cr
& \Delta x.m\Delta v = \frac{h}{{4\pi }},\,\,\Delta x = \frac{h}{{4\pi m\Delta v}} \cr
& {\text{Here}}\,\,\Delta v = \frac{{600 \times 0.005}}{{100}} = 0.03 \cr
& {\text{So}},\,\,\Delta x = \frac{{6.6 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 0.03}} \cr
& = 1.92 \times {10^{ - 3}}{\text{meter}} \cr} $$