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Chemistry · Structure of Atom

JEE Main 2024 — 5 April, Shift 1 — Question 77

The value of Rydberg constant RH=2.18×10−18 JR_H = 2.18 \times 10^{-18}\,\mathrm{J}. The velocity of an electron of mass 9.1×10−31 kg9.1 \times 10^{-31}\,\mathrm{kg} in Bohr's first orbit of hydrogen atom is ____×105 m s−1\_\_\_\_\times 10^{5}\,\mathrm{m\,s^{-1}} (nearest integer).

Answer: 22

Numerical answer — enter this value.

Step-by-step solution

In Bohr orbit, we have

Kinetic energy, K=RHK = R_H

But, K=12mv2K = \frac{1}{2}mv^2

So,

12mv2=2.18×10−18\frac{1}{2}mv^2 = 2.18 \times 10^{-18}

v2=2×2.18×10−189.1×10−31v^2 = \frac{2 \times 2.18 \times 10^{-18}} {9.1 \times 10^{-31}}

v2=4.79×1012v^2 = 4.79 \times 10^{12}

v=2.19×106 m s−1v = 2.19 \times 10^{6}\,\mathrm{m\,s^{-1}} =21.9×105ms−1= 21.9 \times 10^{5}\mathrm {ms^{-1}} Thus, the correct answer is 2222 in nearest integer.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 1
Subject
Chemistry
Chapter
Structure of Atom
Topic
Bohr's Model of Atom

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