Question

If the radius of the first Bohr orbit is $r$, then the de Broglie wavelength of the electron in the $4^{\text{th}}$ orbit will be

• A. $4\pi r$
• B. $6\pi r$
• C. $8\pi r$
• D. $\dfrac{\pi r}{4}$

Hint:

Physics Tip: In Bohr orbit: circumference $=$ integer multiple of wavelength.

Answer 1

The Correct Option is C

Step 1: Use Bohr standing wave condition. According to de-Broglie interpretation of Bohr model: $$ 2\pi r_n=n\lambda_n $$ where
• $r_n$ = radius of $n^{th}$ orbit
• $\lambda_n$ = de-Broglie wavelength

Step 2: Radius of $n^{th$ orbit.} For hydrogen atom: $$ r_n=n^2r_1 $$ Given first orbit radius: $$ r_1=r $$ Hence for fourth orbit: $$ r_4=4^2r=16r $$

Step 3: Apply wave condition for $n=4$. $$ 2\pi r_4=4\lambda_4 $$ $$ \lambda_4=\frac{2\pi(16r)}{4} $$

Step 4: Simplify. $$ \lambda_4=8\pi r $$

Step 5: Conclusion. $$ \therefore \text{Correct option is (C) } 8\pi r. $$