JEE Main · 2009 · 12 Apr 2009 · Offline · Atomic Structure

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 1 0^{- 34} kg m^{2} s^{- 1}

, mass of electron,

e_{m} = 9.1 \times 1 0^{- 31} kg

)

Answer: (C)

Solution

Heisenberg's principle sets a floor on the product of the uncertainties,

Δ x \cdot m Δ v \geq \frac{h}{4 π}

. The speed is known to

0.005

per cent, so

Δ v = 600 \times \frac{0.005}{100} = 0.03 m/s

. The least uncertainty in position is then

Δ x = \frac{h}{4 π m Δ v} = \frac{6.6 \times 1 0 ^{- 34}}{4 \times 3.14 \times 9.1 \times 1 0 ^{- 31} \times 0.03}

, which is about

1.92 \times 1 0^{- 3}

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option