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

We wish to determine the frequency of the de-Broglie wave of the electron in Bohr’s first orbit of hydrogen. In de-Broglie’s theory, the wavelength is given by $$\lambda = \frac{h}{mv}$$, which yields a frequency $$\nu = \frac{v}{\lambda} = \frac{mv^2}{h}\,.$$ Noting that $$mv^2 = 2\times KE$$ and that the kinetic energy in the first orbit is $$|E_1| = R_H = 2.18\times10^{-18}$$ J, we have:

$$\nu = \frac{mv^2}{h} = \frac{2\times KE}{h} = \frac{2\times 2.18\times10^{-18}}{6.6\times10^{-34}}$$

$$= \frac{4.36\times10^{-18}}{6.6\times10^{-34}} = 6.606\times10^{15}\text{ Hz}$$

Expressed in the form $$\_\_\_ \times 10^{13}$$ Hz, this becomes $$\nu = 660.6\times10^{13}\approx661\times10^{13}$$ Hz. The answer is 661.