Question

In the hydrogen atom, the electron makes a transition from the higher orbit ($i$) to a lower orbit ($f$). The ratio of the radius of the orbits in given by $r_i : r_f = 16 : 4$. The wavelength of photon emitted due to this transition is _________ nm. (Given Rydberg constant $= 1.097 \times 10^7 \text{ /m}$)

• A. 121
• B. 242
• C. 486
• D. 974

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The topic of this question is Atomic Physics, focusing on Bohr's model of the hydrogen atom.
We need to use the given radius ratio to identify the quantum numbers of the initial and final states, then calculate the emission wavelength using the Rydberg formula.
Step 2: Key Formula or Approach:
1. Orbit radius relationship: $r_n \propto n^2$.
2. Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Step 3: Detailed Explanation:
The ratio of the radii is:
\[ \frac{r_i}{r_f} = \frac{n_i^2}{n_f^2} = \frac{16}{4} = 4 \]
Taking the square root gives $\frac{n_i}{n_f} = 2$.
Since the ratio is $16:4$, the orbit numbers must be $n_i = 4$ and $n_f = 2$.
Now, we apply the Rydberg formula for a transition from $n = 4 \to 2$ (Balmer series):
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \]
\[ \lambda = \frac{16}{3R} = \frac{16}{3 \times 1.097 \times 10^7} \text{ m} \]
\[ \lambda \approx 4.861 \times 10^{-7} \text{ m} = 486.1 \text{ nm} \]
Step 4: Final Answer:
The wavelength is approximately 486 nm.