the ratio of the radii of n = 10 orbit of hydrogen and `Li^(+2)` ion is .....

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

`r_(n)=(n^(2))/(Z).a_(0)` where `a_(0)=` Bohr radius `=(h^(2)in_(0))/(pie^(2)m)=0.53Å`
For H atom Z=1,n=10
`:.r_(10)=(100)/(1)a_(0)` and For `Li^(+2)` atom Z=3, n=10, r=R
`:.R_(10)=(100)/(3)a_(0)" ":.(r_(10))/(R_(10))-3`