Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is \(5.7 \text{ × }10^5ms{-1}\).
Answer & explanation
Correct answer: option 2
The correct answer is option 2. \(\pm 10^{-10}m\).
Here, \(\Delta v\text{ = }5.7 \text{ × }10^5ms^{-1}\)
\(m\text{ = 9.1}\text{ × }10^{-31}kg\)
\(\Delta x\text{ = }\frac{h}{4\pi \text{ × }m\text{ × }\Delta v}\)
\(\Delta x\text{ = }\frac{6.6\text{ × }10^{-34}kgm^2s^{-1}}{4\text{ × }\frac{22}{7}\text{ × }9.1\text{ × }10^{-31}kg\text{ × }5.7\text{ × }10^5ms{-1}}\)
\(\Delta x\text{ = }1.0 \text{ × }10^{-10}m\)
\(\text{i.e., Uncertainty in the position =}\pm 10^{-10}m\)