The triply ionized beryllium (Be^{3+}) has th | vedclass

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(C) The radius of an electron in a hydrogen-like atom is given by the formula $r_n = a_0 \frac{n^2}{Z}$, where $a_0$ is the Bohr radius, $n$ is the pr

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$n = 2$

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(C) The radius of an electron in a hydrogen-like atom is given by the formula $r_n = a_0 \frac{n^2}{Z}$, where $a_0$ is the Bohr radius, $n$ is the principal quantum number, and $Z$ is the atomic number.For the ground state of hydrogen, $n_H = 1$ and $Z_H = 1$. Thus, $r_H = a_0 \frac{1^2}{1} = a_0$.For triply ionized beryllium $(Be^{3+})$, $Z_{Be} = 4$. Let the energy state be $n_{Be}$. The radius is $r_{Be} = a_0 \frac{n_{Be}^2}{4}$.Given that $r_{Be} = r_H$, we have $a_0 \frac{n_{Be}^2}{4} = a_0$.This simplifies to $\frac{n_{Be}^2}{4} = 1$, which means $n_{Be}^2 = 4$.Therefore, $n_{Be} = 2$.

The triply ionized beryllium $(Be^{3+})$ has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given $Z = 4$ for beryllium)

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