Question

Show that the circumference of the orbit of an electron revolving in the \( n \)-th orbit is equal to \( n\lambda \) with the help of Bohr's quantum theory. Also, show the emission and absorption spectral lines between energy levels \( n = 1 \) and \( n = 3 \) of hydrogen atom. 
 

Hint:

The Bohr model combines classical circular orbits with quantum conditions. Emission corresponds to the release of energy, while absorption involves gaining energy.

Answer 1

1. Bohr's Quantum Theory:
- Bohr's quantization condition states that the angular momentum \( mvr \) of an electron in the \( n \)-th orbit is given by: \[ mvr = \frac{nh}{2\pi}, \quad \text{where } n = 1, 2, 3, \dots \] Here, \( m \) is the mass of the electron, \( v \) is its velocity, \( r \) is the orbital radius, \( h \) is Planck's constant, and \( n \) is the quantum number. 2. Relation with De Broglie Wavelength:
- De Broglie wavelength of the electron is given by \( \lambda = \frac{h}{mv} \).
- Substituting \( \lambda \) in the above expression: \[ 2\pi r = n\lambda. \] Thus, the circumference of the orbit is \( n \lambda \). 3. Emission and Absorption Lines:
- The energy difference between two energy levels \( E_n = -\frac{13.6}{n^2} \, \text{eV} \) is emitted or absorbed as a photon: \[ E = E_f - E_i = -\frac{13.6}{n_f^2} + \frac{13.6}{n_i^2}. \] - For transitions between \( n = 1 \), \( n = 2 \), and \( n = 3 \), the emission and absorption lines correspond to the Lyman and Balmer series in the spectrum.