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 nearlyNEET 2015 (Phase 1)

We know, the radius of the nucleus is R = R 0 m a s s N u m b e r 1 3 R = R 0 A 1 3 R A l R T e = 27 725 1 3 = 3 5 R T e = 5 3 R A l

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PhysicsModern PhysicsNuclear physicsEasy2 minPYQ_2015

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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 nearlyNEET 2015 (Phase 1)

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Solution:

We know, the radius of the nucleus is $R = R_{0} {(m a s s N u m b e r)}^{\frac{1}{3}}$
$R = R_{0} A^{\frac{1}{3}}$
$\frac{R_{A l}}{R_{T e}} = {(\frac{27}{725})}^{\frac{1}{3}} = \frac{3}{5}$
$R_{T e} = \frac{5}{3} R_{A l}$

Stream:NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: PYQ_2015

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