JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Atomic Structure

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from

n = 4

n = 2

He^{+}

spectrum

Answer: (A)

Solution

For a one-electron species

\frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})

. For

He X^{+}

(

Z = 2

) going from

n = 4

n = 2

\frac{1}{λ} = 4 R (\frac{1}{4} - \frac{1}{16}) = \frac{3 R}{4}

. For hydrogen (

Z = 1

) the same wavenumber needs

\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}} = \frac{3}{4}

, satisfied by

n_{1} = 1, n_{2} = 2

. The matching hydrogen transition is from

n = 2

n = 1

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option