Q138 / 180 · COMEDK 2019 · Physics

Which state of triply ionised Beryllium $\left(\mathrm{Be}^{+++}\right)$ has the same orbital radius as that of the ground state of hydrogen?

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

Solution

Radius of $n$ th-orbit in hydrogen like atom is given by $r=\frac{n^{2} h^{2}}{4 \pi^{2} m k Z e^{2}}$ i.e., $\quad r \propto \frac{n^{2}}{Z}$ For hydrogen, $Z=1, n=1$ in ground state $\Rightarrow \quad \frac{n^{2}}{Z}=\frac{1^{2}}{1}=1$ For Beryllium $\left(\mathrm{Be}^{+++}\right), Z=4$ orbital is same $\therefore \quad \frac{n^{2}}{Z}=1$ $\Rightarrow \quad n^{2}=1 \times Z \Rightarrow n^{2}=4$ $\Rightarrow \quad n=2$ Thus, the second level of triply ionised $\mathrm{Be}^{+++}$has same radius as the ground state of hydrogen.

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

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