What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum lines?

(Take $h=6.63×10^{-34} Js$; $c=3×10^8 ms^{-1}$; $e=1.6×10^{-19} C$)

122 nm

The correct answer is Option (1) → 122 nm

The wavelength (λ) of Rydberg formula,

$\frac{1}{λ}=R\left(\frac{1}{{n_i}^2}-\frac{1}{{n_f}^2}\right)$

R = Rydberg Constant = $1.097×10^7m^{-1}$

when, $n_1=1,n_2=2$  [The least energy photon in lyman series is emitted from $n_2=2$ and $n_1=1$]

$∴\frac{1}{λ}=1.097×10^7\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

$≃122nm$