Question

Highly energetic electrons are bombarded on a target of an element containing $30$ neutrons. The ratio of radii of nucleus of that of Helium necleus is $(14{)}^{1/3}$. Find atomic number of the nucleus.
$(R=1.1\times {10}^7{m}^{-1}$ and $c=3\times {10}^{8}m/s)$.

Answer 1

Equation for radius of nucleus of the atomic number is

$R=R_0{A}^{\frac{1}{3}}$, ($\because R_0=1.2fm$)

So, let the atomic number of the atom is A.

Then the radius of the atom is $R=R_O{A}^{\frac{1}{3}}$

& the radius of the helium atom is

$R_H=R_o{4}^{\frac{1}{3}}$

given that the ratio of the radius of nucleus and the radius of helium atom is

$\frac{R_H}{R_A}={14}^{\frac{1}{3}}$

$\therefore \frac{R_A}{R_H}={\left(14\right)}^{\frac{1}{3}}=>\frac{R_O{\left(A\right)}^{\frac{1}{3}}}{R_O{4}^{\frac{1}{3}}}={\left(14\right)}^{\frac{1}{3}}=>{\left(\frac{A}{4}\right)}^{\frac{1}{3}}={\left(14\right)}^{\frac{1}{3}}$

$=>\frac{A}{4}=14,=>A=56$