Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral lin

Q: Why did we convert the wavelength to centimeters instead of meters?

The Rydberg constant ($R_H$) was given in $\text{cm}^{-1}$. To ensure dimensional consistency across the equation, it's easiest to convert the wavelength to centimeters. You could also convert $R_H$ to $\text{m}^{-1}$, which would work just as well!

Q: Why didn't we use Planck's constant ($h$) and the speed of light ($c$)?

The problem provided those values as a distractor! While you *could* calculate the energy $\Delta E = \frac{hc}{\lambda}$ and equate it to the energy difference between Bohr orbits, using the Rydberg formula directly for wavelength is a much faster and more elegant path.

Q: How do I know which $n$ is $n_1$ and which is $n_2$?

In the emission formula $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, $n_1$ is the lower energy level (final state) and $n_2$ is the higher energy level (initial state). This ensures the wavelength is a positive value.

Q: What does $n = \infty$ physically mean?

$n = \infty$ represents an electron that is completely free from the nucleus's attractive pull. A transition from $\infty$ to a lower orbit means a free electron is captured by the atom, emitting a photon in the process.

Q: Why is the Paschen series associated with $n_1 = 3$?

The spectral series are named based on their final destination orbit. Lyman is $n_1=1$, Balmer is $n_1=2$, Paschen is $n_1=3$, Brackett is $n_1=4$, and Pfund is $n_1=5$. It's a standard convention you should memorize!

Question

Heat treatment of muscular pain involves radiation of wavelength of about $900	ext{ nm}$. Which spectral line of H-atom is suitable for this purpose? [$R_H = 1 	imes 10^5	ext{ cm}^{-1}$, $h = 6.6 	imes 10^{-34}	ext{ Js}, c = 3 	imes 10^8	ext{ ms}^{-1}$]

Options:

(A) Paschen, $5 ightarrow 3$

(B) Paschen, $\infty ightarrow 3$

(C) Lyman, $\infty ightarrow 1$

(D) Balmer, $\infty ightarrow 2$

Sigma Percentile · Accepted Answer

### The Healing Power of Light

Imagine an electron in a hydrogen atom jumping down from a higher energy level to a lower one. When it does, it emits a photon—a tiny packet of light. In this problem, we are looking for a specific jump that releases a photon with a wavelength of $900	ext{ nm}$, which happens to be in the infrared region and is used for treating muscular pain.

Our goal is to identify which of the given transitions corresponds to this exact wavelength. Let's look at the possible transitions given in the options and find the perfect match.

### The Master Equation: Rydberg's Formula

To connect the wavelength of the emitted photon to the energy levels, we use the famous Rydberg formula. This elegant equation tells us exactly how the wavelength $\lambda$ relates to the initial and final orbits:

$$\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$$

For a hydrogen atom, the atomic number $Z$ is simply $1$. The problem gives us the Rydberg constant $R_H = 1 	imes 10^5	ext{ cm}^{-1}$.

Now, let's carefully substitute the values we know. The wavelength is $900	ext{ nm}$. Since our Rydberg constant is given in inverse centimeters, we must convert the wavelength to centimeters to maintain dimensional consistency.

$$900	ext{ nm} = 900 	imes 10^{-7}	ext{ cm} = 9 	imes 10^{-5}	ext{ cm}$$

Plugging this and the Rydberg constant into our formula gives us the raw setup:

$$\frac{1}{9 	imes 10^{-5}} = 10^5 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$$

### The Elegant Cancellation

Let's simplify this equation. On the left side, $\frac{1}{9 	imes 10^{-5}}$ can be rewritten as $\frac{10^5}{9}$.

Notice how beautifully the $10^5$ term appears on both sides!

$$\frac{10^5}{9} = 10^5 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$$

We can simply cancel out $10^5$ from both sides. We are left with a very simple and clean relation:

$$\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{9}$$

### Decoding the Transitions

Now we just need to find which of our given transitions satisfies this condition. Let's test option (b), which is a transition from infinity to the third orbit, part of the Paschen series.

Here, the final state is $n_1 = 3$ and the initial state is $n_2 = \infty$. Substituting these into our bracketed term, we get:

$$\frac{1}{3^2} - \frac{1}{\infty^2}$$

Since $\frac{1}{\infty}$ is zero, this perfectly evaluates to:

$$\frac{1}{9} - 0 = \frac{1}{9}$$

### The Final Verdict

And there we have it! The transition from infinity to the third energy level perfectly matches our required condition. This means the correct spectral line suitable for the heat treatment is indeed the Paschen series transition from $\infty ightarrow 3$.

**A Quick Tip:** Did you notice that the problem provided values for Planck's constant ($h$) and the speed of light ($c$)? We didn't even use them! Sometimes, questions provide extra information to test if you can identify the most direct path to the solution. Always trust your core concepts and look for the most elegant method.

Animated Solution for Chemistry - Atomic Structure: Heat treatment of muscular pain involves radiation of wavelength of about $900 nm$ . Which spectral line of H-atom is suitable for this purpose? [ $R_{H} = 1 \times 1 0^{5} cm^{- 1}$ , $h = 6.6 \times 1 0^{- 34} Js, c = 3 \times 1 0^{8} ms^{- 1}$ ]

Select Answer:

Visualized Solution

$Energy Levels of Hydrogen Atom$

$Rydberg Formula$

$Substituting the Values$

$Simplifying the Equation$

$Checking the Options$

$Conclusion$

$Beyond the Problem$

The Sigma Insight: Bohr's Model

The Healing Power of Light

The Master Equation

The Elegant Cancellation

Decoding the Transitions

The Final Verdict

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