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 $= 1.097 \times 10^7 \text{ /m}$)
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 $= 1.097 \times 10^7 \text{ /m}$)
Updated On: Apr 12, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We need to identify the initial and final orbit numbers ($n$) using the radius ratio and then calculate the emission wavelength using the Rydberg formula.
Step 2: Key Formula or Approach:
1. Radius $r_n \propto n^2$.
2. Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Step 3: Detailed Explanation:
Ratio of radii: $\frac{r_i}{r_f} = \frac{n_i^2}{n_f^2} = \frac{16}{4} = 4$.
Taking the square root: $\frac{n_i}{n_f} = 2$.
The transition is to a lower orbit, so possible values could be $n_i = 4, n_f = 2$ or $n_i = 2, n_f = 1$.
From the ratio $16:4$, the most direct interpretation is $n_i=4$ and $n_f=2$.
Now apply the Rydberg formula for $n_i = 4 \to n_f = 2$ (Balmer series):
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16} \]
\[ \lambda = \frac{16}{3R} = \frac{16}{3 \times 1.097 \times 10^7} \text{ m} \]
\[ \lambda \approx \frac{16}{3.291 \times 10^7} \approx 4.861 \times 10^{-7} \text{ m} = 486.1 \text{ nm} \]
Step 4: Final Answer:
The wavelength is approximately 486 nm.
We need to identify the initial and final orbit numbers ($n$) using the radius ratio and then calculate the emission wavelength using the Rydberg formula.
Step 2: Key Formula or Approach:
1. Radius $r_n \propto n^2$.
2. Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Step 3: Detailed Explanation:
Ratio of radii: $\frac{r_i}{r_f} = \frac{n_i^2}{n_f^2} = \frac{16}{4} = 4$.
Taking the square root: $\frac{n_i}{n_f} = 2$.
The transition is to a lower orbit, so possible values could be $n_i = 4, n_f = 2$ or $n_i = 2, n_f = 1$.
From the ratio $16:4$, the most direct interpretation is $n_i=4$ and $n_f=2$.
Now apply the Rydberg formula for $n_i = 4 \to n_f = 2$ (Balmer series):
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16} \]
\[ \lambda = \frac{16}{3R} = \frac{16}{3 \times 1.097 \times 10^7} \text{ m} \]
\[ \lambda \approx \frac{16}{3.291 \times 10^7} \approx 4.861 \times 10^{-7} \text{ m} = 486.1 \text{ nm} \]
Step 4: Final Answer:
The wavelength is approximately 486 nm.
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