If radius of the A 13 27 l nucleus is taken to be R A l then the radius of T e 53 125 nucleus is nearly

If radius of the

{}_{13}^{27}A l

nucleus is taken to be

R_{A l}

then the radius of

{}_{53}^{125}{T e}

nucleus is nearly

${(\frac{53}{13})}^{\frac{1}{3}} R_{A l}$
$\frac{5}{3} R_{A l}$
$\frac{3}{5} R_{A l}$
${(\frac{13}{53})}^{\frac{1}{3}} R_{A l}$

Solution

We know, the radius of the nucleus is

R = R_{0} {(m a s s N u m b e r)}^{\frac{1}{3}}

R = R_{0} A^{\frac{1}{3}}

\frac{R_{A l}}{R_{T e}} = {(\frac{27}{725})}^{\frac{1}{3}} = \frac{3}{5}

R_{T e} = \frac{5}{3} R_{A l}

Asked in: NEET 2015 (Phase 1)

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