The shortest wavelength of H atom in the Lyman series is $\lambda_{1}$. The longest wavelength in the Balmer Series of $\mathrm{He}^{+}$is:

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Atomic Structure

The shortest wavelength of H atom in the Lyman series is

λ_{1}

. The longest wavelength in the Balmer Series of

He^{+}

is:

Answer: (C)

The shortest wavelength of the hydrogen Lyman series comes from the

n = \infty \to 1

transition,

\frac{1}{λ _{1}} = R_{H}

, so

λ_{1} = \frac{1}{R _{H}}

. For the

H e^{+}

ion (nuclear charge

Z = 2

) the longest Balmer wavelength comes from

n = 3 \to 2

, where

\frac{1}{λ} = R_{H} Z^{2} (\frac{1}{4} - \frac{1}{9}) = 4 R_{H} \cdot \frac{5}{36} = \frac{5 R _{H}}{9}

. Hence

λ = \frac{9}{5 R _{H}} = \frac{9 λ _{1}}{5}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option