The radius of the smallest electron orbitin hydrogen like ion is `(0.51xx10^(-10)//4)m`, then it is

Options

A. Hydrogen atom
B. `He^(+)`
C. `Li^(2+)`
D. `Be^(3+)`

Answer

D

Explanation

For hydrogen like atom, the radius of nth orbit
`r_(n)=(n^(2))/(Z)=a_(0)`
here, `a_(0)=0.51xx10^(-10)m`
`therefore r_(n)=(0.51xx10^(-10))/(4)m`
In the ground state, `n=1`
`therefore (0.51xx10^(-16))/(4)=(1^(2))/(Z)xx0.51xx10^(-10)" "therefore Z=4`