Question

The ratio of the de Broglie wavelengths associated with the electron revolving in the first and third orbits in hydrogen atom is :

• A. 2
• B. \( \frac{1}{2} \)
• C. 3
• D. \( \frac{1}{3} \)

Hint:

Memorizing proportionalities in Bohr's model saves massive amounts of exam time:
Radius: \( r \propto n^2 \)
Velocity: \( v \propto 1/n \)
Energy: \( E \propto 1/n^2 \)
Wavelength: \( \lambda \propto n \)

Answer 1

The Correct Option is D

Concept:
• The de Broglie wavelength \( \lambda \) of a moving particle is inversely proportional to its momentum \( p = mv \), given by the equation \( \lambda = \frac{h}{mv} \).

• For an electron in a hydrogen atom, Niels Bohr postulated that its orbital angular momentum must be quantized: \( mvr = \frac{nh}{2\pi} \), where \( n \) is the principal quantum number.

• By rearranging this quantization condition, we can relate the electron's momentum directly to its orbit radius and quantum number: \( mv = \frac{nh}{2\pi r} \).

• Additionally, Bohr's model proves that the radius \( r \) of the \( n^{\text{th}} \) stable orbit is directly proportional to the square of the principal quantum number: \( r \propto n^2 \).

Step 1: Derive the proportionality between wavelength and quantum number
Start with the de Broglie wavelength formula and substitute momentum from Bohr's condition:
\[ \lambda = \frac{h}{mv} \]
Substitute \( mv = \frac{nh}{2\pi r} \):
\[ \lambda = \frac{h}{\left( \frac{nh}{2\pi r} \right)} \]
\[ \lambda = \frac{2\pi r}{n} \]
Now, apply the known fact that orbit radius scales with the square of the quantum number (\( r \propto n^2 \)):
\[ \lambda \propto \frac{n^2}{n} \]
Simplify the expression to reveal a direct linear relationship:
\[ \lambda \propto n \]
This indicates that the de Broglie wavelength of an orbiting electron is simply directly proportional to the principal quantum number of its orbit.

Step 2: Calculate the required ratio
We are asked to find the ratio of the wavelength in the first orbit (\( n_1 = 1 \)) to the wavelength in the third orbit (\( n_3 = 3 \)).
Using the direct proportionality \( \lambda \propto n \):
\[ \frac{\lambda_1}{\lambda_3} = \frac{n_1}{n_3} \]
Substitute the respective orbit numbers:
\[ \frac{\lambda_1}{\lambda_3} = \frac{1}{3} \]

Step 3: Conclusion
The calculated ratio of the de Broglie wavelengths is \( 1/3 \).
This corresponds exactly to option (D).