Question

If the circumference of the nth orbit of an electron is \(S\) and the corresponding de Broglie wavelength of the orbit is \(\lambda\), then on the basis of Bohr's atom model, prove that \(S = n\lambda\).

Hint:

The relationship \(S = n\lambda\) is a result of Bohr’s quantization condition and de Broglie’s wave-particle duality, which ties the electron's orbit and wavelength together.

Answer 1

According to Bohr's atomic model, the angular momentum of an electron in orbit \(n\) is quantized and given by:
\[ L = n \hbar \] Where: - \(L\) is the angular momentum, - \(n\) is the principal quantum number, - \(\hbar = \frac{h}{2\pi}\) is the reduced Planck's constant, where \(h\) is Planck's constant.
The angular momentum is also related to the linear momentum of the electron and its radius \(r\) as:
\[ L = mvr \] Where: - \(m\) is the mass of the electron, - \(v\) is the linear velocity of the electron, - \(r\) is the radius of the orbit.
By equating these two expressions for \(L\):
\[ mvr = n \hbar \] Now, using the de Broglie hypothesis, the wavelength \(\lambda\) of the electron is related to its momentum \(p = mv\) as:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \] Thus, the radius \(r\) of the nth orbit can be expressed as:
\[ r = \frac{n \hbar}{mv} \] The circumference \(S\) of the nth orbit is:
\[ S = 2\pi r \] Substituting the expression for \(r\) from above:
\[ S = 2\pi \times \frac{n \hbar}{mv} \] Now, from the de Broglie relation, we can substitute \(mv = \frac{h}{\lambda}\) into the equation:
\[ S = 2\pi \times \frac{n \hbar}{\frac{h}{\lambda}} = 2\pi \times n\lambda \] Thus, we obtain the desired result:
\[ S = n\lambda \]