Question

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

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

Answer 1

Answer: (B)

n = 2

Answer 2

Radius of nth orbit in hydrogen like atom is $r_{n} = \frac{a_{0}n^{2}}{Z}$

Where $a_{0}$ is the Bohr’s radius

For hydrogen atom Z = 1

$\therefore r_{1} = a_{0}$ ($\because$n = 1 for ground state)

For $Be^{3 +},Z = 4$

$\therefore r_{n} = \frac{a_{0}n^{2}}{4}$

According to given problem

$r_{1} = r_{n}$

$a_{0} = \frac{n^{2}a_{0}}{4}$

$n = 2$