Heat treatment of muscular pain involves radiation of wavelength of about 900nm. Which spectral line of H-atom is suitable for this purpose? `[R_H=1xx10^5 cm^(-1), h=6.6xx10^(-34) Js, c=3xx10^8 ms^(-1) ]`

Options

A. Pashcen, `oorarr3`
B. Lyman, `oorarr1`
C. Paschen, `5rarr3`
D. Balmer, `oorarr2`

Explanation

To solve the problem of identifying which spectral line of the hydrogen atom corresponds to a radiation wavelength of approximately 900 nm, we can use the Rydberg formula:

\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \(\lambda\) is the wavelength,
- \(R_H\) is the Rydberg constant (\(1.097 \times 10^5 \, \text{cm}^{-1}\)),
- \(Z\) is the atomic number (for hydrogen, \(Z = 1\)),
- \(n_1\) is the lower energy level,
- \(n_2\) is the higher energy level.

### Step 1: Convert Wavelength to Appropriate Units
Given that the wavelength \(\lambda\) is 900 nm, we convert it to meters:
\[ \lambda = 900 \, \text{nm} = 900 \times 10^{-9} \, \text{m} \]

### Step 2: Convert Wavelength to Centimeters
Since the Rydberg constant is in \(\text{cm}^{-1}\), we convert \(\lambda\) to centimeters:
\[ \lambda = 900 \times 10^{-9} \, \text{m} = 900 \times 10^{-7} \, \text{cm} = 9.00 \times 10^{-5} \, \text{cm} \]

### Step 3: Apply the Rydberg Formula
Substituting the values into the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
\[ \frac{1}{9.00 \times 10^{-5}} = 1.097 \times 10^5 \cdot 1^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 4: Calculate \(\frac{1}{\lambda}\)
Calculating \(\frac{1}{\lambda}\):
\[ \frac{1}{\lambda} = \frac{1}{9.00 \times 10^{-5}} \approx 1.111 \times 10^4 \, \text{cm}^{-1} \]

### Step 5: Set Up the Equation
Now we can set up the equation:
\[ 1.111 \times 10^4 = 1.097 \times 10^5 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step 6: Solve for Energy Levels
Rearranging gives:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1.111 \times 10^4}{1.097 \times 10^5} \]
Calculating the right side:
\[ \frac{1.111 \times 10^4}{1.097 \times 10^5} \approx 0.1013 \]

### Step 7: Identify Possible Values of \(n_1\) and \(n_2\)
Assuming \(n_2 = \infty\) (as the electron transitions from a higher energy level to a lower one), we have:
\[ \frac{1}{n_1^2} = 0.1013 \implies n_1^2 \approx \frac{1}{0.1013} \approx 9.86 \]
Thus, \(n_1 \approx 3\).

### Conclusion
The transition is from \(n_2 = \infty\) to \(n_1 = 3\), which corresponds to the Paschen series.

### Final Answer
The suitable spectral line of the hydrogen atom for the purpose of heat treatment of muscular pain with a wavelength of about 900 nm is from the **Paschen series**.

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