In the Balmer series spectra of hydrogen , there is a line corresponding to wavelength `4344 Å` Calculate the number of highest orbits from electron drops to greater other lines `(R xx c = 3.289 xx 10^(15))`

Explanation

To solve the problem, we need to calculate the number of the highest orbit (n2) from which an electron drops to the n=2 level in the Balmer series of hydrogen, given the wavelength of the emitted light (4344 Å).

### Step-by-Step Solution:

1. **Convert Wavelength to Meters**:
The given wavelength is 4344 Å. We need to convert this to meters:
\[
\text{Wavelength} (\lambda) = 4344 \, \text{Å} = 4344 \times 10^{-10} \, \text{m}
...