1. Goal

We need the hydrogen spectral line whose wavelength is closest to the radiation used in heat treatment:

$\lambda \approx 900\,\text{nm} = 9 \times 10^{-7}\,\text{m} = 9 \times 10^{-5}\,\text{cm}$

For hydrogen spectral lines, use the Rydberg formula:

$\bar\nu = \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \quad n_2>n_1$

Given:

$R_H = 10^5\,\text{cm}^{-1}$

So we evaluate each option.

2. Option A: Lyman series, $\infty \to 1$

For $n_1=1$, $n_2=\infty$:

$\frac{1}{\lambda} = R_H\left(1-0\right)=10^5\,\text{cm}^{-1}$

$\lambda = \frac{1}{10^5} = 10^{-5}\,\text{cm} = 10^{-7}\,\text{m} = 100\,\text{nm}$

This is far from $900\,\text{nm}$.

So, A is not suitable.

3. Option B: Balmer series, $\infty \to 2$

For $n_1=2$, $n_2=\infty$:

$\frac{1}{\lambda} = R_H\left(\frac{1}{2^2}-0\right)=10^5\cdot\frac{1}{4}=2.5\times 10^4\,\text{cm}^{-1}$

$\lambda = \frac{1}{2.5\times 10^4}=4\times 10^{-5}\,\text{cm}=4\times 10^{-7}\,\text{m}=400\,\text{nm}$

This is also not close to $900\,\text{nm}$.

So, B is not suitable.

4. Option C: Paschen series, $5 \to 3$

For $n_1=3$, $n_2=5$:

$\frac{1}{\lambda} = 10^5\left(\frac{1}{3^2}-\frac{1}{5^2}\right)$

$=10^5\left(\frac{1}{9}-\frac{1}{25}\right) = 10^5\left(\frac{25-9}{225}\right)=10^5\cdot\frac{16}{225}$

$\frac{1}{\lambda} \approx 7.11\times 10^3\,\text{cm}^{-1}$

$\lambda \approx \frac{1}{7.11\times 10^3} \approx 1.406\times 10^{-4}\,\text{cm}$

Converting to nm:

$1.406\times 10^{-4}\,\text{cm} = 1.406\times 10^{-6}\,\text{m} = 1406\,\text{nm}$

Not close to $900\,\text{nm}$.

So, C is not suitable.

5. Option D: Paschen series, $\infty \to 3$

For $n_1=3$, $n_2=\infty$:

$\frac{1}{\lambda} = 10^5\left(\frac{1}{9}-0\right)=\frac{10^5}{9}\,\text{cm}^{-1}$

$\lambda = \frac{9}{10^5}\,\text{cm}=9\times 10^{-5}\,\text{cm}$

Convert to meters:

$9\times 10^{-5}\,\text{cm} = 9\times 10^{-7}\,\text{m} = 900\,\text{nm}$

This exactly matches the required wavelength.

So, D is suitable.

6. Final Answer

The spectral line of hydrogen suitable for radiation of wavelength about $900\,\text{nm}$ is:

$\boxed{\text{Paschen series, } \infty \to 3}$

So the correct option is D.