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 : $\mathrm{R}_{\mathrm{H}}$ (Rydberg constant) $=2.18 \times 10^{-18} \mathrm{~J}, h$ (Plank's constant) $=6.6 \times 10^{-34} \mathrm{~J} . \mathrm{s}$.]

<p>$$\begin{aligned} & \lambda=\frac{h}{m v} \\ & \lambda \cdot v=\frac{h}{m} \\ & \frac{\lambda \cdot v^2}{v}=\frac{h}{m} \\ & \frac{v^2}{\text { Frequency }}=\frac{h}{m} \end{aligned}$$</p> <p>$$\begin{aligned} & \text { Frequency }=\frac{\mathrm{mv}^2}{\mathrm{~h}}=\frac{2 \mathrm{R}_{\mathrm{H}}}{\mathrm{h}} \\ & =\frac{2 \times 2.18 \times 10^{-18}}{6.6 \times 10^{-34}} \\ & =660.6 \times 10^{13} \mathrm{~Hz} \end{aligned}$$</p> <p>Nearest integer $=661$</p> <p>If we take $h=6.626 \times 10^{-34}$</p> <p>Then</p> <p>$$\begin{aligned} \text { Frequency } & =\frac{2 R_H}{h} \\ & =658.01 \times 10^{13} \mathrm{~Hz} \end{aligned}$$</p> <p>But if value of $\mathrm{h}$ is given as $6.6 \times 10^{-34}$ correct answer will be 661.</p>