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

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:
$\nu \approx 661 \times 10^{13} \, \text{Hz}$

Answer 2

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:
$\nu \approx 661 \times 10^{13} \, \text{Hz}$