Wave number of the first line in the Balmer series of $\mathrm{Be}^{3+}$ is $2.5 \times 10^5 \mathrm{~cm}^{-1}$. Wave number of the second line of the Paschen series of $\mathrm{Li}^{2+}$ is

$7.2 \times 10^5 \mathrm{~cm}^{-1}$

$1.8 \times 10^4 \mathrm{~cm}^{-1}$

JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

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Application +4 | -1

Wave number of the first line in the Balmer series of

\mathrm{Be}^{3+}

2.5 \times 10^5 \mathrm{~cm}^{-1}

. Wave number of the second line of the Paschen series of

\mathrm{Li}^{2+}

View answer and explanation

Correct option: A

7.2 \times 10^4 \mathrm{~cm}^{-1}

Option A (

7.2 \times 10^4 \mathrm{~cm}^{-1}

) is correct because it is calculated using the Rydberg formula and the given wavenumber of the first line in the Balmer series of

\mathrm{Be}^{3+}

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n^{\text {th }}

\mathrm{A}

\times 10^{12}

\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}}

\ell = 1,\,\,m = 0

1.25 \times {10^{ - 13}} \times {A^{1/3}}\mathrm{cm}.