When the electron orbiting in hydrogen atom goes from one orbit to another orbit (principal quantum number \( = n \)), the de-Broglie wavelength (\( \lambda \)) associated with it is related to \( n \) as}
Show Hint
The circumference of the $n^{th}$ orbit is exactly $n$ de-Broglie wavelengths.
Step 1: Understanding the Concept:
According to de-Broglie's hypothesis, a moving particle is associated with a matter wave.
In Bohr's model of the hydrogen atom, an electron moves in circular orbits. The condition for a stable orbit is that the circumference of the orbit must contain an integral number of de-Broglie wavelengths. Step 2: Key Formula or Approach:
Bohr's quantization condition for angular momentum is:
\[ mvr = \frac{nh}{2\pi} \]
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
We also need the proportionality of the orbit radius \( r \) with the principal quantum number \( n \):
\[ r \propto n^2 \]
Step 3: Detailed Explanation:
From the de-Broglie wavelength equation, we can write the momentum as:
\[ mv = \frac{h}{\lambda} \]
Substitute this expression for \( mv \) into Bohr's quantization condition:
\[ \left(\frac{h}{\lambda}\right) r = \frac{nh}{2\pi} \]
Rearranging this equation to solve for the de-Broglie wavelength \( \lambda \):
\[ \lambda = \frac{2\pi r}{n} \]
We know that in a hydrogen atom, the radius of the \( n \)-th orbit \( r \) is directly proportional to the square of the principal quantum number \( n \):
\[ r = a_0 n^2 \implies r \propto n^2 \]
Substitute this proportionality into the expression for \( \lambda \):
\[ \lambda \propto \frac{2\pi (n^2)}{n} \]
\[ \lambda \propto \frac{n^2}{n} \]
\[ \lambda \propto n \]
Therefore, the de-Broglie wavelength associated with an orbiting electron is directly proportional to the principal quantum number \( n \). Step 4: Final Answer:
The relationship is \( \lambda \propto n \).