Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1152: 10

Answer

(a) $32$ (b) $P^{32}_{15}$

Work Step by Step

(a) We know that $r_1=(1.2\times 10^{-15})A_1^{\frac{1}{3}}$.......eq(1) Where $r_1$ is the radius of the nucleus of an alpha particle and $A_1$ is the mass number. If we double the radius, then the above equation becomes $2r_1=(1.2\times 10^{-15})A_2^{\frac{1}{3}}$......eq(2) Dividing eq(1) by eq(2), we have $\frac{r_1}{2r_1}=\frac{(1.2\times 10^{-15})A_1^{\frac{1}{3}}}{(1.2\times 10^{-15})A_2^{\frac{1}{3}}}$ This simplifies to: $\frac{1}{2}=\frac{4^{\frac{1}{3}}}{A_2^{\frac{1}{3}}}$ $\implies A_2=(2\times 4^{\frac{1}{3}})^3$ $A_2=32$ As the number of nucleons is equal to the mass number of the element, thus the required number of nucleons in the nucleus is $32$. (b) We know that phosphorus has 23 isotopes and they have the same atomic number but different mass number. The mass number of the phosphorus which has a radius twice that of alpha particle is $32$, hence the required symbol can be written as $P^{32}_{15}$.
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