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NEET CHEMISTRYStructure of AtomEasy

Question

Uncertainty in position of an electron (e−e^-) and Helium atom (He) is similar. If uncertainty in momentum of electron is 32×10532 \times 10^5, then uncertainty in momentum of Helium will be:

A

32 \times 10^5

B

16 \times 10^5

C

8 \times 10^5

D

None of the above

Step-by-Step Solution

According to Heisenberg's Uncertainty Principle, the product of uncertainty in position (Δx\Delta x) and uncertainty in momentum (Δp\Delta p) is greater than or equal to a constant (h4π\frac{h}{4\pi}). The mathematical expression is: Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi} Given that the uncertainty in position is similar (equal) for both the electron and the Helium atom (Δxe=ΔxHe\Delta x_e = \Delta x_{He}), the uncertainty in their momentum must also be equal to satisfy the principle, regardless of their masses. ΔpHe=Δpe=32×105\Delta p_{He} = \Delta p_e = 32 \times 10^5 Note: Mass would only result in a difference if the question asked for uncertainty in velocity (Δv\Delta v), since Δp=mΔv\Delta p = m\Delta v.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Structure of Atom. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYStructure of Atomuncertaintypositionelectronheliumsimilar

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