JEE PYQ: Nuclear Physics Question 12
Question 12 - 2020 (03 Sep Shift 2)
The radius $R$ of a nucleus of mass number $A$ can be estimated by the formula $R = (1.3 \times 10^{-15})A^{1/3}$ m. It follows that the mass density of a nucleus is of the order of: ($M_{\text{prot.}} \cong M_{\text{neut.}} = 1.67 \times 10^{-27}$ kg)
(1) $10^3$ kg m$^{-3}$
(2) $10^{10}$ kg m$^{-3}$
(3) $10^{24}$ kg m$^{-3}$
(4) $10^{17}$ kg m$^{-3}$
Show Answer
Answer: (4)
Solution
Density of nucleus, $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{mA}{\frac{4}{3}\pi R^3}$ $\Rightarrow \rho = \frac{mA}{\frac{4}{3}\pi(R_0 A^{1/3})^3}$ ($\because R = R_0 A^{1/3}$) Here $m$ = mass of a nucleon $\therefore \rho = \frac{3 \times 1.67 \times 10^{-27}}{4 \times 3.14 \times (1.3 \times 10^{-15})^3}$ $\Rightarrow \rho = 2.38 \times 10^{17}$ kg/m$^3$
BETA
