Question

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

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

Answer 1

Answer: (B)

n = 2

Answer 2

Radius of n orbit in hydrogen like atoms is $r_{n}=\frac{a_{0}n^{2}}{Z}$ where a is the Bohr?? radius For hydrogen atom, Z = 1 ? r = a (? n = 1 for ground state) For Be, Z = 4 $\therefore\quad r_{n}=\frac{a_{0}n^{2}}{4}$ According to given problem $\quad$r = r $a_{0}=\frac{n^{2}a_{0}}{4}$ n = 2