What is the diameter of `._(2)Be^(4)` in this ground state? Given Bohr radius of hydrogen atom is 53pm.

Options

A. `53p m`
B. `26.5p m`
C. `1.6p m`
D. `100p m`

Answer

A

Explanation

On the basis of Bohr's theory,
`r=(n^(2)h^(2))/(4pi^(2)mKZe^(2))=a_(0) (n^(2))/Z`
where `a_(0)`=radius of hydrogen atom 53pm.
for `._(2)He^(4)`, Z=2 , n=1 for ground state
`:. r=53xx(1^(2))/2=26.5p m`
Diameter`=2r=2xx26.5p m=53p m`