JEE MAIN - Chemistry (2002 - No. 55)
Uncertainty in position of a minute particle of mass 25 g in space is 10-5 m. What is the uncertainty in its velocity (in ms-1) (h = 6.6 $$\times$$ 10-34 Js)
2.4 $$\times$$ 10-34
0.5 $$\times$$ 10-34
2.1 $$\times$$ 10-28
0.5 $$\times$$ 10-23
Explanation
According to Hysenberg's Uncertainty Principal
$$$\Delta x . \Delta p \ge { h \over {4\pi}}$$$
Where $$\Delta x $$ is uncertainty in position and $$\Delta p $$ is uncertainty in momentum and as in momentum m (mass of object) is always same so we can say
$$$\Delta x . m. \Delta v \ge { h \over {4\pi}}$$$
Mass of the particle (m) = 25 g = 0.025 kg
Position of the particle ($$\Delta x$$) = 10-5 m
$$\Delta x . \Delta p $$ = $${ h \over {4\pi}}$$
$$\Delta x . m . \Delta v $$ = $${ h \over {4\pi}}$$
$$\Delta v$$ = $${h \over {4 \times \Delta x \times m \times \pi }}$$
$$\Delta v$$ = $${{6.6 \times {{10}^{ - 34}}} \over {4 \times {{10}^{ - 5}} \times 0.025 \times 3.14}}$$
$$\Delta v$$ = 2.1 $$\times$$ 10-28
Position of the particle ($$\Delta x$$) = 10-5 m
$$\Delta x . \Delta p $$ = $${ h \over {4\pi}}$$
$$\Delta x . m . \Delta v $$ = $${ h \over {4\pi}}$$
$$\Delta v$$ = $${h \over {4 \times \Delta x \times m \times \pi }}$$
$$\Delta v$$ = $${{6.6 \times {{10}^{ - 34}}} \over {4 \times {{10}^{ - 5}} \times 0.025 \times 3.14}}$$
$$\Delta v$$ = 2.1 $$\times$$ 10-28
Comments (0)
