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Friday, 9 October

JEE Main 2024 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 9 April 2024, Shift 2, Chemistry Q81

Question 81 of 90 in this shift, Chemistry question 21 of 30, Section B.

Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter 10−1510^{-15} m is ________ ×109 ms−1\times 10^{9}\ \mathrm{ms^{-1}} (nearest integer) [Given : mass of electron =9.1×10−31= 9.1\times10^{-31} kg, Plank's constant (h)=6.626×10−34(h) = 6.626\times10^{-34} Js] (Value of π=3.14\pi = 3.14)

Official answer

58

NTA final key.

Chapter
Structure of Atom
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.