In an atom, an electron is moving with the speed of 600 m/s have an accuracy of 0.005%. The minimum uncertainity with which the position of the electron can be located is: h=6.6×10−34Js,mass of electron,em=9.1×10−31kg
A
5.10×10−3m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.92×10−3m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3.84×10−3m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.52×10−3m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B1.92×10−3m Given: Speed of electron = 600 m/s Accuracy of speed = 0.005% Uncertainity in speed (Δv)=600×0.005100=0.03m/s Let uncertainity in the position be Δx. According to Heisenberg's uncertainity principle: mΔvΔx≥h4π ∴Minimum Δx=h4πmΔv=6.6×10−344×3.14×9.1×10−31×0.03=1.92×10−3m