Question

Uncertainty in the position of an electron (mass = 9.1 × 10⁻³¹ kg)
moving with a velocity 300 ms⁻¹, accurate upto 0.001% will be
(h = 6.63 × 10⁻³⁴ Js)
(a) 1.92 × 10⁻² m (b) 3.84 × 10⁻² m
(c) 19.2 × 10⁻² m (d) 5.76 × 10⁻² m

Answer 1

answer : option (c) 19.2 × 10^-2 m

given, mass of electron, m = 9.1 × 10^-31 Kg

speed of electron, v = 300 m/s

uncertainty in speed of electron, ∆v = 0.01% of v

= 0.001 × 300/100 = 3 × 10^-3 m/s

using Heisenberg's uncertainty principle,

∆x = h/4πm∆v

= (6.625 × 10^-34 J.s)/(4 × 3.14 × 9.1 × 10^-31 kg × 3 × 10^-3 m/s)

= 6.625 × 10^-34/(12.56 × 27.3 × 10^-34 )

= 0.0192 m

= 19.2 × 10^-2 m

hence option (c) is correct choice.

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Answer 2

Uncertainty in the position of an electron moving with a velocity 300 ms⁻¹, accurate up to 0.001% will be 1.92 × 10⁻² m

Explanation:

According to Heisenberg uncertainty principle, the position and momentum of small particles like electron can not be found instantly.

Mathematically

$\Delta x \cdot \Delta v \geq \frac{h}{4 \pi m}$

Given that

mass $m = 9.1 \times10^{-31} kg$

uncertainty in velocity $\Delta v =300 ms^{-1}$

Form the equation

$\Delta x \geq \frac{h}{4 \pi m\Delta v}$

$\Rightarrow \Delta x=\frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 300 \times 0.001 \times 10^{-2}} \approx 0.0192 =1.92 \times 10^{-2} m$

Therefore Uncertainty in the position = 1.92 × 10⁻² m

Hence, the correct answer is option (a)