Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is \ \ \ \ × 10 13 \ Hz (nearest integer). [Given: R H (Rydberg constant) = 2.18 × 10 -18 \ J , h (Planck's constant) = 6.6 × 10 -34 \ J\,s .]

Chemistry · Structure of Atom

JEE Main 2024 — 6 April, Shift 1 — Question 81

Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is $\_\_\_\_$ $\times 10^{13}\ \mathrm{Hz}$ (nearest integer).

[Given: $\mathrm{R_H}$ (Rydberg constant) $= 2.18 \times 10^{-18}\ \mathrm{J}$ , $\mathrm{h}$ (Planck's constant) $= 6.6 \times 10^{-34}\ \mathrm{J\,s}$ .]

Answer: 661

Numerical answer — enter this value.

Step-by-step solution

de-Broglie relation, $\mathrm{\lambda = \frac{h}{mv}}$

Wave frequency relation, $\mathrm{\nu = \frac{v}{\lambda}}$

On substituting, we have

\mathrm{\nu = \frac{mv^2}{h}}

Given kinetic energy:

\mathrm{\frac{1}{2}mv^2 = 2.18 \times 10^{-18}\ J}

\mathrm{mv^2 = 4.36 \times 10^{-18}\ J}

\mathrm{\nu = \frac{4.36 \times 10^{-18}}{6.6 \times 10^{-34}}}

\mathrm{\nu = 6.606 \times 10^{15}\ Hz}

\mathrm{\nu = 660.6 \times 10^{13}\ Hz}

\mathrm{\nu \approx 661 \times 10^{13}\ Hz}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Chemistry
Chapter: Structure of Atom
Topic: Wave-Particle Duality of Matter - de Broglie, Heisenberg