Q135 / 180 · AIIMS 2014 · Physics

A triply ionized beryllium $\left(\mathrm{Be}^{3+}\right)$ has the same orbital radius as the ground state of hydrogen. Then the quantum state $n$ of $\mathrm{Be}^{3+}$ is

  • A
    $n=1$
  • Correct answer
    B
    $n=2$
  • C
    $n=3$
  • D
    $n=4$

Solution

Radius of $n^{\text {th }}$ orbit in hydrogen like atoms is $r_n=\frac{a_0 n^2}{Z}$ where $a_0$ is the Bohr's radius For hydrogen atom, $Z=1$ $\therefore r_1=a_0 \quad(\because n=1$ for ground state $)$ For $\mathrm{Be}^{3+}, Z=4$ $\therefore r_n=\frac{a_0 n^2}{4}$ According to given problem $r_1=r_n$ $a_0=\frac{n^2 a_0}{4}$ $n=2$

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

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