Question:

The Bohr radius of a hydrogen like species is \(70.53\,\text{pm}\). The species and the stationary state \((n)\) are respectively. (Given: Hydrogen atom Bohr radius is \(52.9\,\text{pm}\)).

Updated On: Apr 12, 2026
  • \(Li^{2+},\,3\)
  • \(He^{+},\,3\)
  • \(He^{+},\,2\)
  • \(Li^{2+},\,2\)
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The Correct Option is B

Solution and Explanation

Concept: Radius of orbit in hydrogen-like species: \[ r_n = a_0 \frac{n^2}{Z} \] where \[ a_0 = 52.9\,\text{pm} \] Step 1: {Substitute given radius.} \[ 70.53 = 52.9 \frac{n^2}{Z} \] \[ \frac{n^2}{Z} = \frac{70.53}{52.9} \] \[ \frac{n^2}{Z} \approx 1.333 \] \[ =\frac{4}{3} \] Step 2: {Check possible species.} For \(He^+\): \[ Z=2 \] \[ \frac{n^2}{2}=\frac{4}{3} \] \[ n^2=\frac{8}{3}\approx2.67 \] Nearest integer \(n=3\). Thus species: \[ He^+,\,n=3 \]
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