The shortest wavelength of the spectral lines emitted in Balmer Series is: [Rydberg constant = $10^7 m^{−1}$]
Answer & explanation
Correct answer: option 3
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, $n_2=∞$
$∴\frac{1}{λ}=R_H\left(\frac{1}{2^2}-\frac{1}{∞^2}\right)$
$=10^7×\frac{1}{4}$
$⇒λ=4×10^{-7}m=4000Å$