Which state of triply ionised Beryllium `(Be^(+++))` the same orbital radius as that of the ground state hydrogen ?

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_(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)=(a_(0)n^(2))/(4)`
According to given problem,
`r_(1)=r_(n), a_(0)=(n^(2)a_(0))/(4) rArr n=2`