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

Answer 2

The problem asks for the frequency of the de-Broglie wave associated with an electron in the first Bohr orbit of a hydrogen atom.

Concept Used:

The term "frequency of the de-Broglie wave" can be interpreted in the context of the Bohr model as the classical frequency of revolution of the electron in its orbit. The de-Broglie wave for a stable orbit forms a standing wave, and its properties are intrinsically linked to the dynamics of the orbiting electron.

The frequency of revolution (\(f\)) for an electron in the n-th Bohr orbit of a hydrogen atom is given by the formula:

\[ f_n = \frac{2 R_H}{h n^3} \]

where:

• \(R_H\) is the Rydberg constant in units of energy (Joules).
• \(h\) is Planck's constant.
• \(n\) is the principal quantum number.

This formula can be derived from the expression for the total energy of the electron, \(E_n = -R_H/n^2\), and classical mechanics relationships for circular motion under a Coulomb force.

Step-by-Step Solution:

Step 1: Identify the given values and the specific orbit.

• Rydberg constant, \(R_H = 2.18 \times 10^{-18} \, \text{J}\)
• Planck's constant, \(h = 6.6 \times 10^{-34} \, \text{J.s}\)
• The electron is in the first Bohr orbit, which means the principal quantum number is \(n = 1\).

Step 2: Substitute the given values into the formula for the frequency of revolution.

We use the formula \(f_n = \frac{2 R_H}{h n^3}\) with \(n=1\):

\[ f_1 = \frac{2 R_H}{h (1)^3} = \frac{2 R_H}{h} \]

Step 3: Perform the calculation.

\[ f_1 = \frac{2 \times (2.18 \times 10^{-18} \, \text{J})}{6.6 \times 10^{-34} \, \text{J.s}} \] \[ f_1 = \frac{4.36 \times 10^{-18}}{6.6 \times 10^{-34}} \, \text{Hz} \] \[ f_1 \approx 0.6606 \times 10^{16} \, \text{Hz} \] \[ f_1 = 6.606 \times 10^{15} \, \text{Hz} \]

Final Computation & Result:

The problem asks for the frequency to be expressed in the format ______ \( \times 10^{13} \, \text{Hz} \).

We need to convert our calculated frequency to this format:

\[ f_1 = 6.606 \times 10^{15} \, \text{Hz} = 660.6 \times 10^{13} \, \text{Hz} \]

Rounding this value to the nearest integer, we get 661.

The frequency of the de-Broglie wave is 661 \( \times 10^{13} \, \text{Hz} \).