Question

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

Answer 1

Correct Answer: 658

The de-Broglie wavelength $\lambda$ is given by:
\[\lambda = \frac{h}{mv}\]
For an electron in motion:
\[\text{Kinetic Energy (K.E.)} = \frac{1}{2} mv^2 \implies v^2 = \frac{2 \cdot \text{K.E.}}{m}.\]
Step 1: Substituting values:
\[\text{K.E.} = R_H = 2.18 \times 10^{-18} \, \text{J}.\]
\[v = \sqrt{\frac{2 \cdot R_H}{m}} = \sqrt{\frac{2 \cdot 2.18 \times 10^{-18}}{9.1 \times 10^{-31}}}.\]
Step 2: Using frequency relation:
\[\nu = \frac{v}{\lambda} = \frac{h}{mv}.\]
Step 3: Substituting $h$ and solving for $\nu$:
\[\nu = \frac{\text{K.E.}}{h} = \frac{2.18 \times 10^{-18}}{6.6 \times 10^{-34}}.\]
\[\nu = 660.6 \times 10^{13} \, \text{Hz}.\]
Step 4: Nearest integer:
\[\nu \approx 661 \times 10^{13} \, \text{Hz}\]