The electron in the $\mathrm{n}^{\text{th}}$ orbit of $\mathrm{Li}^{2+}$ is excited to ($\mathrm{n}+1$) orbit using the radiation of energy $1.47 \times 10^{-17} \mathrm{~J}$ (as shown in the diagram). The value of $n$ is ______. Given $\mathrm{R}_{\mathrm{H}}=2.18 \times 10^{-18} \mathrm{~J}$

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Atomic Structure

The electron in the

n^{th}

orbit of

Li^{2 +}

is excited to (

n + 1

) orbit using the radiation of energy

1.47 \times 1 0^{- 17} J

(as shown in the diagram). The value of

n

is ______.

Given

R_{H} = 2.18 \times 1 0^{- 18} J

Answer: 1

For a hydrogen-like ion the energy absorbed in moving from level

n

n + 1

Δ E = R_{H} Z^{2} (\frac{1}{n ^{2}} - \frac{1}{( n + 1 ) ^{2}})

. For

L i^{2 +}

Z = 3

Z^{2} = 9

. Substituting,

1.47 \times 1 0^{- 17} = 2.18 \times 1 0^{- 18} \times 9 (\frac{1}{n ^{2}} - \frac{1}{( n + 1 ) ^{2}})

, which gives

\frac{1}{n ^{2}} - \frac{1}{( n + 1 ) ^{2}} = \frac{1.47 \times 1 0 ^{- 17}}{1.962 \times 1 0 ^{- 17}} = \frac{3}{4}

. This is met by

n = 1

, since

1 - \frac{1}{4} = \frac{3}{4}

. The value of

n

is 1.

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Integer answer