If the shorts series limit of the balmer series for hydrogen is `3644 Å`, find the atomic number of the element which gives X-ray wavelength down to `1 Å`. Identify the element.

Explanation

To solve the problem, we need to find the atomic number of the element that produces X-ray wavelengths down to 1 Å (angstrom) given that the shortest wavelength of the Balmer series for hydrogen is 3644 Å.

### Step-by-Step Solution:

1. **Understanding the Shortest Wavelength of the Balmer Series**:
The shortest wavelength of the Balmer series corresponds to the transition from n=2 to n=∞. The formula for the wavelength in the Balmer series is given by:
\[
\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
...