QJEEM24Ch8478: JEE Main 2024 Chemistry - Class 11 JEEM24Ch8478 Solution: Frequency of the de-Broglie wave of electron in Bohr's first…

$\begin{aligned}& \lambda=\frac{h}{m v} \\& \lambda \cdot v=\frac{h}{m} \\& \frac{\lambda \cdot v^2}{v}=\frac{h}{m} \\& \frac{v^2}{\text { Frequency }}=\frac{h}{m}\end{aligned}$

$\begin{aligned}& \text { Frequency }=\frac{\mathrm{mv}^2}{\mathrm{~h}}=\frac{2 \mathrm{R}_{\mathrm{H}}}{\mathrm{h}} \\& =\frac{2 \times 2.18 \times 10^{-18}}{6.6 \times 10^{-34}} \\& =660.6 \times 10^{13} \mathrm{~Hz}\end{aligned}$

Nearest integer $=661$

If we take $h=6.626 \times 10^{-34}$

Then

$\begin{aligned}\text { Frequency } & =\frac{2 R_H}{h} \\& =658.01 \times 10^{13} \mathrm{~Hz}\end{aligned}$

But if value of $\mathrm{h}$ is given as $6.6 \times 10^{-34}$ correct answer will be 661.