A measurement of an electron's speed is $2.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ and has an uncertainty of $10 \%$. What is the minimum uncertainty in the position of the electron? $\left( m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , h = 6.626 \times \right.$ $\left. 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } \right)$
A) $4.9 \mathrm {~nm}$
B) $0.58 \mathrm {~nm}$
C) $1.5 \mathrm {~nm}$
D) $0.87 \mathrm {~nm}$
E) $0.29 \mathrm {~nm}$

Question

Quizplus · Accepted Answer

The minimum uncertainty in the position ($\Delta x$) of the electron can be calculated using the Heisenberg Uncertainty Principle, which is given by $\Delta x \Delta p \geq \frac{h}{4\pi}$, where $\Delta p$ is the uncertainty in momentum and $h$ is Planck's constant. The uncertainty in the electron's speed ($\Delta v$) is $10\%$ of $2.0 \times 10^6 \, \text{m/s}$, so $\Delta v = 0.1 \times 2.0 \times 10^6 \, \text{m/s} = 2.0 \times 10^5 \, \text{m/s}$. The uncertainty in momentum ($\Delta p$) is given by $\Delta p = m_{\text{electron}} \Delta v = 9.11 \times 10^{-31} \, \text{kg} \times 2.0 \times 10^5 \, \text{m/s}$. Substituting the values into the Heisenberg Uncertainty Principle formula and solving for $\Delta x$ gives a value close to $0.58 \, \text{nm}$.

A Measurement of an Electron's Speed Is 2.0×106 m/s2.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }2.0×106 m/s

A Measurement of an Electron's Speed Is $2.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$