Question

Given that the mass of electron is $9.1\times {10}^{-31}$ kg and velocity of the electron is $2.2\times {10}^{8}m/s$ uncertainly in its velocity is 0.1% the uncertainly in position would be

• A. 26nm
• B. 32nm
• C. 48nm
• D. 50nm

Answer 1

Answer: (B)

32nm

$\begin{array}{c}\text{According to Hisenberg uncertainity rule}\\ \Delta p\Delta x=\frac{h}{4\pi }\end{array}$

$\begin{array}{c}△p\text{is uncertainity in momentum(mv)}\\ \Delta x\text{is uncertainity in position}\\ h\text{is planck's constant.}\end{array}$

$\begin{array}{cc}\therefore m\Delta V& \Delta x=\frac{h}{4\pi }\\ \text{Given}& m=9.1\times {10}^{-31}kg\\ & \frac{\Delta V}{V}\%=0.1\%\\ & h=6.626\times {10}^{-34}Js\\ & V=2.2\times {10}^8\\ & \Delta V=2.2\times {10}^5\end{array}$

$\begin{array}{cc}\therefore \Delta x& =\frac{6.626\times {10}^{-34}}{9.1\times {10}^{-31}\times 4\pi \times 2.2\times 105}\\ \Delta x& =32nm\end{array}$