Which state of the triply ionized Beryllium `(Be^(3+))` has the same orbit radius as that of the ground state of hydrogen atom?

Options

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

Answer

B

Explanation

Radius of `n^(th)` orbit in hydrogen like atoms is `r_n=(a_0n^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=4therefore r_n=(a_0n^2)/(4)`
According to given problem
`r_1=r_nrArra_0=(n^2a_0)/(4)rArrn=2`