Minimum uncertainty in speed of an electron | JEE Main 2022 | Chemca Solution

Chemca.in

Class 11 Class 12

1. Home
2. Structure of Atom
3. JEE Numerical Practice

Numerical Type JEE Main 2022 (29 Jul Shift-1) +4 / -1 Marking

The minimum uncertainty in the speed of an electron in a one-dimensional region of length \( 2a_0 \) (Where \( a_0 \) = Bohr radius = 52.9 pm) is _________ \(\text{km s}^{-1}\). (Nearest integer)

Given Data:

• Mass of electron (\(m\)) = \( 9.1 \times 10^{-31} \text{ kg} \)
• Planck's constant (\(h\)) = \( 6.63 \times 10^{-34} \text{ J s} \)

Enter your answer (integer value in km/s):

Detailed Step-by-Step Solution

This problem is a direct application of Heisenberg's Uncertainty Principle. The key is identifying the correct maximum uncertainty in position (\( \Delta x \)) to find the minimum uncertainty in velocity (\( \Delta v \)).

Step 1: Identify the Uncertainty in Position (\( \Delta x \))

The electron is confined to a one-dimensional region of length \( 2a_0 \). The maximum uncertainty in locating the electron is the length of this region itself.

\( \Delta x = 2a_0 = 2 \times 52.9 \text{ pm} = 105.8 \text{ pm} \)

Convert picometers to meters for standard SI calculations:

\( \Delta x = 105.8 \times 10^{-12} \text{ m} \)

Step 2: Apply Heisenberg's Formula

According to the principle:

\( \Delta x \cdot \Delta p \ge \frac{h}{4\pi} \)

Since \( \Delta p = m \cdot \Delta v \), we can substitute and rearrange for \( \Delta v \):

\( \Delta v = \frac{h}{4\pi \cdot m \cdot \Delta x} \)

Step 3: Substitute the Values

\( \Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.1415 \times (9.1 \times 10^{-31}) \times (105.8 \times 10^{-12})} \)

First, calculate the denominator carefully:

\( \text{Denominator} = (4 \times 3.1415 \times 9.1 \times 105.8) \times 10^{-43} \)

\( \text{Denominator} \approx 12097.5 \times 10^{-43} \)

Now calculate \( \Delta v \) in m/s:

\( \Delta v = \frac{6.63 \times 10^{-34}}{12097.5 \times 10^{-43}} \)

\( \Delta v = \left( \frac{6.63}{12097.5} \right) \times 10^9 \)

\( \Delta v \approx 0.0005480 \times 10^9 \text{ m/s} = 548000 \text{ m/s} \)

Step 4: Convert to required units (km/s)

The question specifically asks for the answer in \( \text{km s}^{-1} \). We know \( 1 \text{ km} = 1000 \text{ m} \):

\( \Delta v = \frac{548000}{1000} \text{ km/s} = 548 \text{ km/s} \)

Conclusion: The calculated minimum uncertainty in velocity is exactly 548 km/s. The integer value is 548.

Avoiding Common Pitfalls in JEE Numericals

Questions on Heisenberg's Uncertainty Principle are frequent in JEE Main. A major trap here is forgetting to convert the final answer from \( \text{m/s} \) to \( \text{km/s} \). Precision with powers of 10 is crucial. To master these calculations and understand the theory behind quantum mechanical models, read our in-depth article on the Structure of Atom Class 11 Chemistry.

Enhance Your Chemistry Preparation

• Explore comprehensive revision notes, solved examples, and NCERT solutions for Class XI Chemistry.
• Preparing for your Class 12 Boards and competitive exams simultaneously? Accelerate your prep with our advanced Class XII Chemistry resources.

© 2026 Chemca.in. All rights reserved. Providing top-tier JEE and NEET solutions for Indian students.

🏠 Home 📝 Quiz 📘 Revision Notes ⬇ Downloads 📚 Class Notes ▶ Video Lectures

Get new posts by email

Powered by

📚 Also Read

Lecture Notes