Solveeit Logo
Login

Question

Chemistry Question on Structure of atom

Given, the mass of electron is 9.11×10−31 kg9.11 \times 10^{-31}\, kg, Planck's constant is 6.626×10−34 Js6.626 \times 10^{-34}\, Js, the uncertainty involved in the measurement of velocity within a distance of 0.1A˚0.1 \mathring{A} is

A

5.79×106ms−15.79 \times 10^6 ms^{-1}

B

5.79×107ms−15.79 \times 10^7 ms^{-1}

C

5.79×108ms−15.79 \times 10^8 ms^{-1}

D

5.79×105ms−15.79 \times 10^5 ms^{-1}

Answer

5.79×106ms−15.79 \times 10^6 ms^{-1}

Explanation

Solution

By Heisenberg's uncertainty principle

15mmΔx×Δpx≥h4π15mm \Delta x \times \Delta px \ge \frac{h}{4\pi}
or10 mmΔx×(mvx)≥h4πor 10\,mm \Delta x \times (mv_x) \ge \frac{h}{4\pi}
15mmΔx×Δvx≥h4πm15mm \Delta x\times \Delta v_x \ge \frac{h}{4\pi m}
Δp\Delta p= uncertainty in momentum

Δx \Delta x=uncertainty in position

Δv \Delta v=uncertainty in velocity

m = mass of particle

Given that,

Δx=0.1A˚=0.1×10−10m\Delta x = 0.1 \mathring{A}= 0.1 \times 10^{-10} m
m=9.11×10−31kgm = 9.11 \times 10^{-31} kg

h = Planck's constant = 6.626×10−346.626 \times 10^{-34}

n = 3.14

Thus,

Δv×0.1×10−10=6.626×10−344×3.14×9.11×10−31\Delta v\times 0.1 \times 10^{-10}=\frac{6.626 \times 10^{-34}}{4\times3.14 \times9.11\times10^{-31}}
Δv=6.626×10−344×3.14×9.11×10−31×0.1×10−10ms−1\Delta v=\frac{6.626 \times 10^{-34}}{4\times3.14 \times 9.11 \times 10^{-31}\times0.1 \times10^{-10}}ms^{-1}
=5.785×106ms−1=5.79×106ms−1=5.785 \times 10^6 ms^{-1}=5.79\times 10^6 ms^{-1}