Modern Physics 5 Question 12
12. Order of magnitude of density of uranium nucleus is $\left(m _p=1.67 \times 10^{-27} kg\right)$
$(1999,2 M)$
(a) $10^{20} kg / m^{3}$
(b) $10^{17} kg / m^{3}$
(c) $10^{14} kg / m^{3}$
(d) $10^{11} kg / m^{3}$
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Answer:
Correct Answer: 12. (d)
Solution:
- Radius of a nucleus is given by
$$ \begin{aligned} R & =R _0 A^{1 / 3} \quad\left(\text { where } R _0=1.25 \times 10^{-15} m\right) \\ & =1.25 A^{1 / 3} \times 10^{-15} m \end{aligned} $$
Here $A$ is the mass number and mass of the uranium nucleus will be
$$ \begin{aligned} & m \approx A m _p \quad \text { where } \quad m _p=\text { mass of proton } \\ &=A\left(1.67 \times 10^{-27} kg\right) \\ & \therefore \text { Density } \rho=\frac{\text { mass }}{\text { volume }}=\frac{m}{\frac{4}{3} \pi R^{3}} \\ &=\frac{A\left(1.67 \times 10^{-27} kg\right)}{A\left(1.25 \times 10^{-15} m\right)^{3}} \text { or } \rho \approx 2.0 \times 10^{17} kg / m^{3} \end{aligned} $$
