The shortest wavelength of the spectral lines emitted in Balmer Series is: [Rydberg constant = $10^7 m^{−1}$]

4000Å

The correct answer is Option (3) → 4000Å

As per Balmer series, ($n_1=2$)

$\frac{1}{λ}=R_H\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$

and, for shortest wavelength, $n_2=∞$

$∴\frac{1}{λ}=R_H\left(\frac{1}{2^2}-\frac{1}{∞^2}\right)$

$=10^7×\frac{1}{4}$

$⇒λ=4×10^{-7}m=4000Å$