Order of magnitude of density of uranium nucleus is ({mₚ} = 1.67 × {10^{ - 27}}kg)

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13.Nuclei

hard

Order of magnitude of density of uranium nucleus is $({m_p} = 1.67 \times {10^{ - 27}}kg)$

A

${10^{20}}kg/{m^3}$

B

${10^{17}}kg/{m^3}$

C

${10^{14}}kg/{m^3}$

D

${10^{11}}kg/{m^3}$

(IIT-1999)

Solution

(b) The order of magnitude of mass and volume of uranium nucleus will be

$m \simeq \, A(1.67 \times 10^{-{27}}\, kg)$ ($A$ is atomic number)

$V = \frac{4}{3}\pi {r^3}\tilde – \frac{4}{3}\pi \;{[(1.25 \times {10^{ – 15}}m){A^{1/3}}]^3}$

$\simeq – (8.2 \times {10^{ – 45}}{m^3})A$

Hence, $\rho = \frac{m}{V} = \frac{{A(1.67 \times {{10}^{ – 27}}kg)}}{{(8.2 \times {{10}^{ – 45}}{m^3})A}}$

$\simeq – 2.0 \times {10^{17}}kg/{m^3}$.

Standard 12

Physics

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