Atoms & Nuclei

JEE Main level, single correct, medium

167 of 239

Question 167 of 239

The mass number of nucleus having radius equal to half of the radius of nucleus with mass number 192192 is:

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Answer: (A) 2424

Given: R1=R22{R}_{1}=\dfrac{{R}_{2}}{2}
As we know, size of the nucleus is given by R=R0(A)1/3R={R}_{0}{\left(A\right)}^{1/3}
Therefore, we can write R0(A1)1/3=R02(A2)1/3{R}_{0}{\left({A}_{1}\right)}^{1/3}=\dfrac{{R}_{0}}{2}{\left({A}_{2}\right)}^{1/3}
⇒A1=18A2\Rightarrow {A}_{1}=\dfrac{1}{8}{A}_{2}
⇒A1=1928=24\Rightarrow {A}_{1}=\dfrac{192}{8}=24

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