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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2024 · 6 April · Morning Shift · Chemistry · Structure of Atom

Frequency of the de-Broglie wave of electron in Bohr's first orbit...

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 : RH( Rydberg constant )=2.18×10−18J,h (Plank’s  constant )=6.6×10−34J.s.] \begin{aligned} & \text{ [Given : } \mathrm{R}_{\mathrm{H}}(\text{ Rydberg constant })=2.18 \times 10^{-18} \mathrm{ J}, h \text{ (Plank's } \\ & \text{ constant })=6.6 \times 10^{-34} \mathrm{ J} \text{.s.] } \end{aligned})$

Correct answer

0

Step-by-step solution

λ=hmvλ=hvmv2mv2h=vλ=v (frequency)  Given 12mv2=2.18×10−18Jh=6.6×10−34v=4.36×10−186.6×10−34=660.60×1013Hz≈661×1013Hz\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{mv}} \\ & \lambda=\frac{\mathrm{hv}}{\mathrm{mv}^2} \\ & \frac{\mathrm{mv}^2}{\mathrm{ h}}=\frac{\mathrm{v}}{\lambda}=\mathrm{v} \text{ (frequency) } \\ & \text{ Given } \frac{1}{2} \mathrm{mv}^2=2.18 \times 10^{-18} \mathrm{ J} \\ & \mathrm{ h}=6.6 \times 10^{-34} \\ & \mathrm{v}=\frac{4.36 \times 10^{-18}}{6.6 \times 10^{-34}}=660.60 \times 10^{13} \mathrm{ Hz} \\ & \approx 661 \times 10^{13} \mathrm{ Hz}\end{aligned}
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