Question:
In an atom, electron is moving with a speed of $x~ms^{-1}.$ If its speed is measured within an accuracy of 0.001%, what is its uncertainty in position (in m)? $(m_{e}=9\times10^{-31}kg, h=6.6\times10^{-34}Js)$
In an atom, electron is moving with a speed of $x~ms^{-1}.$ If its speed is measured within an accuracy of 0.001%, what is its uncertainty in position (in m)? $(m_{e}=9\times10^{-31}kg, h=6.6\times10^{-34}Js)$
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Always ensure all units are in the MKS system (meters, kilograms, seconds) before applying the Heisenberg formula to avoid errors in the power of 10. The accuracy percentage must always be converted to a decimal factor (0.001% = 0.00001) before multiplying by the speed.
Updated On: Jun 8, 2026
- $\frac{3\pi x}{55}$
- $\frac{55\pi}{3x}$
- $\frac{55}{3\pi x}$
- $\frac{55x}{3\pi}$
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The Correct Option is C
Solution and Explanation
Concept: This problem is a direct application of the Heisenberg Uncertainty Principle, which states that it is impossible to simultaneously determine both the exact position and the exact momentum of a subatomic particle. The mathematical expression is given by:
\[
\Delta x \cdot \Delta p \geq \frac{h}{4\pi}
\]
where $\Delta x$ is the uncertainty in position, $\Delta p$ is the uncertainty in momentum, and $h$ is Planck's constant. Since $\Delta p = m \Delta v$, the inequality can be rewritten as:
\[
\Delta x \geq \frac{h}{4\pi m \Delta v}
\]
Step 1: Calculate the uncertainty in velocity ($\Delta v$).
The speed of the electron is given as $x~ms^{-1}$. The uncertainty is measured within an accuracy of 0.001%. \[ \Delta v = x \times \left( \frac{0.001}{100} \right) = x \times 10^{-5} \text{ m/s} \]
Step 2: Substitute the given values into the Heisenberg equation.
Given $m_e = 9 \times 10^{-31} \text{ kg}$, $h = 6.6 \times 10^{-34} \text{ Js}$, and $\Delta v = x \times 10^{-5} \text{ m/s}$: \[ \Delta x \geq \frac{6.6 \times 10^{-34}}{4 \times \pi \times (9 \times 10^{-31}) \times (x \times 10^{-5})} \]
Step 3: Simplify the mathematical expression to find $\Delta x$.
First, multiply the constants in the denominator: \[ 4 \times 9 \times 10^{-31} \times 10^{-5} = 36 \times 10^{-36} \] Now, place this back into the fraction: \[ \Delta x \geq \frac{6.6 \times 10^{-34}}{36 \times \pi \times 10^{-36} \times x} \] Rearranging the powers of 10: \[ \Delta x \geq \frac{6.6 \times 10^2}{36 \times \pi \times x} = \frac{660}{36 \pi x} \] Reducing the fraction $\frac{660}{36}$ by dividing both numerator and denominator by 12: \[ \Delta x \geq \frac{55}{3 \pi x} \]
Step 1: Calculate the uncertainty in velocity ($\Delta v$).
The speed of the electron is given as $x~ms^{-1}$. The uncertainty is measured within an accuracy of 0.001%. \[ \Delta v = x \times \left( \frac{0.001}{100} \right) = x \times 10^{-5} \text{ m/s} \]
Step 2: Substitute the given values into the Heisenberg equation.
Given $m_e = 9 \times 10^{-31} \text{ kg}$, $h = 6.6 \times 10^{-34} \text{ Js}$, and $\Delta v = x \times 10^{-5} \text{ m/s}$: \[ \Delta x \geq \frac{6.6 \times 10^{-34}}{4 \times \pi \times (9 \times 10^{-31}) \times (x \times 10^{-5})} \]
Step 3: Simplify the mathematical expression to find $\Delta x$.
First, multiply the constants in the denominator: \[ 4 \times 9 \times 10^{-31} \times 10^{-5} = 36 \times 10^{-36} \] Now, place this back into the fraction: \[ \Delta x \geq \frac{6.6 \times 10^{-34}}{36 \times \pi \times 10^{-36} \times x} \] Rearranging the powers of 10: \[ \Delta x \geq \frac{6.6 \times 10^2}{36 \times \pi \times x} = \frac{660}{36 \pi x} \] Reducing the fraction $\frac{660}{36}$ by dividing both numerator and denominator by 12: \[ \Delta x \geq \frac{55}{3 \pi x} \]
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