Q2 / 60 · KVPY 2014 (SA) · Chemistry

If the radius of the hydrogen atom is $53 \mathrm{pm}$, the radius of the $\mathrm{He}^{+}$ion is closest to

  • A
    $108 \mathrm{pm}$
  • B
    $81 \mathrm{pm}$
  • Correct answer
    C
    $27 \mathrm{pm}$
  • D
    $13 \mathrm{pm}$

Solution

$r_{n}=\frac{R_{H} n^{2}}{Z}$ $r_{H_{e+}}=\frac{53 n^{2}}{Z}$ $=\frac{53 \times 1^{2}}{2}=27$ approx.

Hence, the correct answer is option: 'C'.

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