If Bohr's theory is applicable to `._(100)Fm^(257)`, then radius of this atom in Bohr's unit is :

Options

A. 100
B. 200
C. 4
D. `1//4`

Answer

D

Explanation

`(r_(m))=(m^(2)/z) (0.53 Å)=(nxx0.53)Å :. m^(2)/z=n`
`m=5` for `._(100)Fm^(257)` (the outermost shell) and `z=100`
`:. N=((5)^(2))/(100)=1/4`