Question
Given, the mass of electron is $$9.11 \times {10^{ - 31}}kg,$$ Planck’s constant is $$6.626 \times {10^{ - 34}}Js,$$ the uncertainty involved in the measurement of velocity within a distance of $$0.1\mathop {\text{A}}\limits^{\text{o}} $$ is
A.
$$5.79 \times {10^6}m{s^{ - 1}}$$
B.
$$5.79 \times {10^7}m{s^{ - 1}}$$
C.
$$5.79 \times {10^8}m{s^{ - 1}}$$
D.
$$5.79 \times {10^5}m{s^{ - 1}}$$
Answer :
$$5.79 \times {10^6}m{s^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{By Heisenberg's}}\,\,{\text{uncertainty principle}} \cr
& \Delta x \times \Delta {p_x} \geqslant \frac{h}{{4\pi }} \cr
& {\text{or}}\,\,\Delta x \times \Delta \left( {m{v_x}} \right) \geqslant \frac{h}{{4\pi }} \cr
& \Delta x \times \Delta {v_x} \geqslant \frac{h}{{4\pi m}} \cr
& \Delta p = {\text{uncertainty in momentum}} \cr
& \Delta x = {\text{uncertainty in position}} \cr
& \Delta v = {\text{uncertainty in velocity}} \cr
& m = {\text{mass of particle}} \cr
& {\text{Given that,}} \cr
& \Delta x = 0.1\mathop {\text{A}}\limits^{\text{o}} = 0.1 \times {10^{ - 10}}m \cr
& m = 9.11 \times {10^{ - 31}}kg \cr} $$
$$h = {\text{Planck's constant}}$$ $$ = 6.626 \times {10^{ - 34}}Js$$
$$\eqalign{
& \pi = 3.14 \cr
& {\text{Thus,}} \cr} $$
$$\Delta v \times 0.1 \times {10^{ - 10}}$$ $$ = \frac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 9.11 \times {{10}^{ - 31}}}}$$
$$\Delta v$$ $$ = \frac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 9.11 \times {{10}^{ - 31}} \times 0.1 \times {{10}^{ - 10}}}}m{s^{ - 1}}$$
$$\eqalign{
& \,\,\,\,\,\,\, = 5.785 \times {10^6}m{s^{ - 1}} \cr
& \,\,\,\,\,\,\, = 5.79 \times {10^6}m{s^{ - 1}} \cr} $$