Ionisation energy of {He}^{+}is 19.6 × 10^{-18} \ {J} \ {atom}^{-1}. The energy of the first stationary state ( n = 1) of {Li}²+} is [AIEEE 2010]

Correct Answer is : −4.41×10−17-4.41 × 10^{-17}−4.41×10−17 J atom−1^{-1}−1

Correct Answer is : −4.41×10−17-4.41 × 10^{-17}−4.41×10−17 J atom−1^{-1}−1

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AIEEE 2010 Solved Paper

Section: Chemistry

Question : 2 of 89

Marks: +1, -0

Ionisation energy of

\mathrm{He}^{+}

19.6 \times 10^{-18} \ \mathrm{J} \ \mathrm{atom}^{-1}

( n =

\mathrm{Li}^{2+}

Solution:

Ionisation energy(IE) - It is the energy required to move an electron from ground state to infinity.

\text{IE}=E_{\infty}-E_1=0-E_1=-E_1

\therefore E_1

\mathrm{He}^{+} = -19.6 \times 10^{-18} \ \mathrm{J} \ \mathrm{atom}^{-1}

Energy of a species at

n

state,

\; (E_n)_{\text{species}} = (E_n)_{\text{hydrogen}} \times Z^2

\; \therefore (E_1)_{\text{hydrogen}} = \frac{-19.6 \times 10^{-18}}{4} [\text{ For He, } Z=2]

\; (E_1)_{L^{+}+2} = \frac{-19.6 \times 10^{-18}}{4} \times 3^2

= -4.41 \times 10^{-17} \ \mathrm{J} \ \mathrm{atom}^{-1}

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