Question

A triply ionized beryllium ($B{e}^{+++}$) has the same orbital radius as the ground state of hydrogen. Then the quantum state $n$ of $B{e}^{3+}$ is

• A. n = 1
• B. n = 2
• C. n = 3
• D. n = 4

Answer 1

Answer: (B)

Radius of n^th orbit in hydrogen like atoms is
$r_n=\frac{a_0{n}^2}{Z}$
where a₀ is the Bohr’s radius
For hydrogen atom, Z = 1
∴ r₁ = a₀ (∵ n = 1 for ground state)
For Be³⁺, Z = 4
$\therefore r_n=\frac{a_0{n}^2}{4}$
According to given problem
r₁ = r_n
$a_0=\frac{n^2{a}_0}{4}$
n = 2