Question

Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of the nucleus to that of the helium nucleus is (14)1/3. The atomic number of the nucleus will be

Answer 1

Answer:

The atomic number of the nucleus will be 26.

Explanation:

Radius of nucleus of an element = R

Atomic mass = Number of neutron + Number of protons

$R=r_o\times A^{\frac{1}{3}}$

$R=r_o\times (Z+30)^{\frac{1}{3}}$..(1)

Radius of nucleus of helium = R'

$R'=r_o\times (4)^{\frac{1}{3}}$..(2)

Dividing (1)and (2) .

$\frac{R}{R'}=14^{\frac{1}{3}}=\frac{r_o\times (Z+30)^{\frac{1}{3}}}{r_o\times (4)^{\frac{1}{3}}}$

Z = 26

The atomic number of the nucleus will be 26.