Question

Order of magnitude of density of uranium nucleus is $\left[ {{m_p} = 1.67 \times {{10}^{ - 27}}kg} \right]$
A. $4 \times {10^{20}}kg\,{m^{ - 3}}$0
B. $1.8 \times {10^{17}}kg\,{m^{ - 3}}$
C. $4.2 \times {10^{14}}kg\,{m^{ - 3}}$
D. $1.4 \times {10^{11}}kg\,{m^{ - 3}}$

Answer 1

The order of magnitude for a given quantity is used to make approximate comparisons. We can calculate the density of the uranium nucleus using the formula as density is equal to mass divided by volume. Mass of uranium will be the mass of the proton multiplied by the atomic mass number.

Complete step by step answer:
To calculate the density of the uranium nucleus, we know that density is mass divided by volume. Let us first find the mass of the uranium nucleus and then we will calculate the volume of the nucleus of a uranium. The mass of the uranium nucleus is the sum of masses of protons and neutrons in the nucleus. This sum is known as atomic mass number and it is denoted as A. Therefore, the total mass of the nucleus will be:
$m = A \times {m_p}$
$\Rightarrow m = A \times 1.67 \times {10^{ - 27}}$ --equation $1$
Radius of the nucleus is given as:
$R = {R_0}{A^{\dfrac{1}{3}}}$
Where ${R_0} = 1.25 \times {10^{ - 15}}m$
${R_0} \simeq 1.3 \times {10^{ - 15}}m$
Volume of the spherical nucleus will be given as:
$V = \dfrac{4}{3}\pi {R^3}$
$\Rightarrow V = \dfrac{4}{3}\pi {\left( {1.3 \times {{10}^{ - 15}}{A^{\dfrac{1}{3}}}} \right)^3}$
$\Rightarrow V = \dfrac{4}{3}\pi A{\left( {1.3 \times {{10}^{ - 15}}} \right)^3}$ --equation $2$
Now, we have the mass of the nucleus and volume of the nucleus. Therefore, the density $D$ will be given as:
$D = \dfrac{m}{V}$
Substituting the values from both equations, we have:
$D = \dfrac{{A \times 1.67 \times {{10}^{ - 27}}}}{{\dfrac{4}{3}\pi A{{\left( {1.3 \times {{10}^{ - 15}}} \right)}^3}}}$
$\Rightarrow D = \dfrac{{3 \times 1.67 \times {{10}^{ - 27}}}}{{4\left( {3.14} \right){{\left( {1.3 \times {{10}^{ - 15}}} \right)}^3}}}$
On solving this we get,
$\therefore D = 1.8 \times {10^{17}}kg\,{m^{ - 3}}$
Therefore, the density of uranium nucleus is $1.8 \times {10^{17}}kg\,{m^{ - 3}}$ .Clearly the magnitude of order is ${10^{17}}$ .

Hence,option B is the correct option.

Note: While calculating the mass of the nucleus, take the total mass of protons and neutrons present inside the nucleus. Since electrons are considered as massless when compared with the mass of proton and neutron. For this reason, it is said that all the mass of an atom resides at the centre of the atom inside the nucleus.