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Structure of Atom question

2010 · Shift 0 · Q25
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Structure of Atom question

2010 · Shift 0 · Q25

JEE MainChemistryStructure of AtomMCQ+4 / −1
Ionisation energy of He+He^+He+ is 19.6 x 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+Li^{2+}Li2+ is
  1. A
    4.41 x 10–16 J atom–1
  2. B
    -4.41 x 10–17 J atom–1
  3. C
    -2.2 x 10–15 J atom–1
  4. D
    8.82 x 10–17 J atom–1
View written solutionFree

Correct answer: B

  1. Use the hydrogen-like ion energy relation

For a hydrogen-like species, En=−2.18×10−18Z2n2  J atom−1E_n = -2.18 \times 10^{-18} \frac{Z^2}{n^2}\; \text{J atom}^{-1}En​=−2.18×10−18n2Z2​J atom−1

Also, the ionisation energy from the ground state is equal to the magnitude of the ground-state energy: I.E.=∣E1∣\text{I.E.} = |E_1|I.E.=∣E1​∣

  1. Given data for He+He^+He+

For He+He^+He+, Z=2Z=2Z=2 and its ionisation energy is given as 19.6×10−18  J atom−119.6 \times 10^{-18}\; \text{J atom}^{-1}19.6×10−18J atom−1

So, E1(He+)=−19.6×10−18  J atom−1E_1(He^+) = -19.6 \times 10^{-18}\; \text{J atom}^{-1}E1​(He+)=−19.6×10−18J atom−1

  1. Relate Li2+Li^{2+}Li2+ to He+He^+He+

For hydrogen-like ions, energy varies as Z2Z^2Z2.

  • For He+He^+He+: Z=2Z=2Z=2
  • For Li2+Li^{2+}Li2+: Z=3Z=3Z=3

Hence, E1(Li2+)E1(He+)=3222=94\frac{E_1(Li^{2+})}{E_1(He^+)} = \frac{3^2}{2^2} = \frac{9}{4}E1​(He+)E1​(Li2+)​=2232​=49​

Therefore, E1(Li2+)=−(19.6×10−18)×94E_1(Li^{2+}) = -\left(19.6 \times 10^{-18}\right)\times \frac{9}{4}E1​(Li2+)=−(19.6×10−18)×49​

  1. Calculate

19.6×94=19.6×2.25=44.119.6 \times \frac{9}{4} = 19.6 \times 2.25 = 44.119.6×49​=19.6×2.25=44.1

So, E1(Li2+)=−44.1×10−18E_1(Li^{2+}) = -44.1 \times 10^{-18}E1​(Li2+)=−44.1×10−18 =−4.41×10−17  J atom−1= -4.41 \times 10^{-17}\; \text{J atom}^{-1}=−4.41×10−17J atom−1

  1. Match with options

This corresponds to:

Option B: −4.41×10−17  J atom−1-4.41 \times 10^{-17}\; \text{J atom}^{-1}−4.41×10−17J atom−1

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