Highly energetic electrons are bombarded on a target of an element containing `30` neutrons. The ratio of radii of nucleus to that of Helium nucleus is `141//3`. The atomic number of nucleus will be.

Options

A. 25
B. 26
C. 56
D. 30

Answer

B

Explanation

(b) By using `R = R_0 A^(1//3) rArr (R_1)/(R_2) = ((A_1)/(A_2))^(1//3)`
`rArr (R)/(R_(He)) = ((A)/(4))^(1//3) rArr (14)^(1//3) = ((A)/(4))^(1//3)`
`rArr A = 56` so `Z = 56 - 30 = 26`.