Ionisation energy of $\ce{He+}$ is $19.6 \times 10^{-18} \text{ J atom}^{-1}$. The energy of the first stationary state $(n=1)$ of $\ce{Li^{2+}}$ is

(B) $-4.41 \times 10^{-17} \text{ J atom}^{-1}$

(A) $4.41 \times 10^{-16} \text{ J atom}^{-1}$

(C) $-2.2 \times 10^{-15} \text{ J atom}^{-1}$

(D) $8.82 \times 10^{-17} \text{ J atom}^{-1}$

JEE Main · 2010 · 11 Apr 2010 · Offline · Atomic Structure

Ionisation energy of

He X^{+}

19.6 \times 1 0^{- 18} J atom^{- 1}

. The energy of the first stationary state

(n = 1)

Li X^{2 +}

Answer: (B)

For a one-electron ion the ground-state energy scales as

Z^{2}

. The ionisation energy of

H e^{+}

(

Z = 2

) is the magnitude of its

n = 1

energy,

19.6 \times 1 0^{- 18}

J. For

L i^{2 +}

(

Z = 3

) the magnitude is larger by the factor

\frac{3 ^{2}}{2 ^{2}} = \frac{9}{4}

, giving

19.6 \times 1 0^{- 18} \times \frac{9}{4} = 4.41 \times 1 0^{- 17}

J. A bound state has negative energy, so

E_{1} (L i^{2 +}) = - 4.41 \times 1 0^{- 17}

J atom

^{- 1}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option