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Structure of Atom question

2025 · 22 Jan · Shift 1 · Q18
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Structure of Atom question

2025 · 22 Jan · Shift 1 · Q18

JEE MainChemistryStructure of AtomMCQ+4 / −1
Radius of the first excited state of Helium ion is given as : a0→\mathrm{a}_0 \rightarrowa0​→ radius of first stationary state of hydrogen atom.
  1. A
    r=a04\mathrm{r}=\frac{\mathrm{a}_0}{4}r=4a0​​
  2. B
    r=2a0\mathrm{r}=2 \mathrm{a}_0r=2a0​
  3. C
    r=4a0\mathrm{r}=4 \mathrm{a}_0r=4a0​
  4. D
    r=a02\mathrm{r=\frac{a_0}{2}}r=2a0​​
View written solutionFree

Correct answer: B

  1. Use Bohr's radius formula for hydrogen-like species

For a hydrogen-like ion,

rn=a0n2Zr_n = a_0\frac{n^2}{Z}rn​=a0​Zn2​

where:

  • a0a_0a0​ = radius of first Bohr orbit of hydrogen,
  • nnn = principal quantum number,
  • ZZZ = atomic number.
  1. Identify the species

Helium ion means He+\mathrm{He}^+He+, which is a hydrogen-like species with:

Z=2Z=2Z=2
  1. First excited state

The first excited state corresponds to:

n=2n=2n=2
  1. Substitute into the formula
r2=a0222=a042=2a0r_2 = a_0\frac{2^2}{2} = a_0\frac{4}{2} = 2a_0r2​=a0​222​=a0​24​=2a0​
  1. Match with the options

Thus,

r=2a0r = 2a_0r=2a0​

So the correct option is B.

  1. Compare with stored correct answer

Stored correct answer: B

This matches our derived answer.

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