Chemistry, 25.10.2019 08:43 christianconklin22

How much energy must be supplied to break a single 21ne nucleus into separated protons and neutrons if the nucleus has a mass of 20.98846 amu?

Answers: 3

Show answers

Answers

Answer from: AbbypiePink4942

Binding energy for a single $^{21}{\text{Ne}}$ nucleus is $\boxed{2.6937\; \times {{10}^{ - 11}}\;{\text{J}}}$ .

Further Explanation:

The smallest particle that is the building block of matter is known as an atom. Most of the space in an atom is empty and its mass is concentrated inside a small region called the nucleus.

Protons, electrons, and neutrons are the three subatomic particles that are present in the atom. The sum of protons and neutrons are called nucleons because these are located in the nucleus of the atom. Electrons revolve around the nucleus in fixed orbital path. The positively charged subatomic particle is called proton. Neutron is a subatomic particle that is present in nucleus with no charge. Electron is the negatively charged subatomic particle. The number of electrons is equal to the number of protons in a neutral atom.

An atom is written as $_{\text{Z}}^{\text{A}}{\text{X}}$ , where A is the atomic mass or mass number, Z is the atomic number, and X is the letter symbol of the element.

Atomic number (Z) is equal to the number of protons that are present in the nucleus of the atom. Atomic mass (A) is the total number of nucleons that are present within the atomic nucleus of the atom.

The given atom is $^{21}{\text{Ne}}$ .

According to the periodic table, Ne has atomic number 10 and mass number 21. We can say that Ne atom has 10 protons and 21 nucleons.

The formula to calculate the number of neutrons is as follows.

${\text{number of neutrons}} = {\text{number of nucleons}} - {\text{number of protons}}$ ......(1)

Substitute10 for numbers of protons and 21 for numbers of nucleons in equation (1).

$\begin{aligned}{\text{number of neutrons}}&=21 - 10\\&=11\\\end{aligned}$

Mass of 1 proton is 1.0073 amu.

Therefore the mass of 10 protons can be calculated as follows:

$\begin{aligned}{\text{Mass of 10 protons}}&=10\left( {{\text{mass of}}\;{\text{1}}\;{\text{proton}}} \right)\\ &={\text{10}}\times{\text{1}}{\text{.0073 amu}}\\&={\text{10}}{\text{.073 amu}}\\\end{aligned}$

Mass of 1 neutron is 1.0087 amu.

Therefore the mass of 11 neutrons can be calculated as follows:

$\begin{aligned}{\text{Mass of 11 neutrons}} &=11\left( {{\text{mass of 1 neutron}}} \right)\\&= 11 \times {\text{1}}{\text{.0087 amu}}\\&={\text{11}}{\text{.0957 amu}} \\\end{aligned}$

Total mass of nucleons can be calculated as follows:

$\begin{aligned}{\text{Total mass of nucleons}}&={\text{mass of 10 protons}} + {\text{mass of 11 neutrons}}\\&{\text{ = 10}}{\text{.073 amu}} + 11.0957{\text{amu}}\\&{\text{ = 21}}{\text{.1687 amu}}\\\end{aligned}$

Mass of Ne nucleus is 21.1687 amu.

The formula to calculate the difference in mass is as follows:

$\Delta {\text{m}} = {\text{total mass of nucleons}} - {\text{mass of Ne nucleus}}$ ......(1)

Here, $\Delta {\text{m}}$ is the mass difference.

Substitute 21.1687 amu for mass of nucleons and 20.98846 amu for mass of Ne nucleus in equation (1).

$\begin{aligned}\Delta {\text{m}} &= 21.1687{\text{ amu}} - {\text{20}}{\text{.98846 amu}}\\&={\text{0}}{\text{.18024 amu}}\\\end{aligned}$

The conversion factor to convert0.18024 amu mass into kg for a single nucleus is as follows:

$\begin{aligned}\Delta {\text{m}}&= 0.18024{\text{ amu}}\left({\frac{{1\;{\text{kg}}}}{{{\text{6}}{\text{.022}}\times {{10}^{26}}{\text{amu}}}}}\right)\\&=2.99302 \times {10^{ - 28}}{\text{ kg}}\\\end{aligned}$

The formula to calculate the binding energy is as follows:

${\text{E}} = \Delta {\text{m}} \times {{\text{c}}^2}$ ......(2)

Here, E is the binding energy of nucleons.

c is the speed of light.

$\Delta {\text{m}}$ is the mass difference.

Substitute $2.99302 \times {10^{ - 28}}{\text{ kg}}$ for $\Delta {\text{m}}$ and $3 \times {10^8}\;{\text{m/s}}$ for c in equation (1).

$\begin{aligned}{\text{E}}&=\left( {2.99302 \times {{10}^{ - 28}}{\text{ kg}}}\right){\left( {3 \times {{10}^8}\;{\text{m/s}}} \right)^2}\\&=\left( {2.6937 \times {{10}^{ - 11}}\;{\text{kg}} \cdot {{\text{m}}^2}{\text{/}}{{\text{s}}^2}}\right)\left( {\frac{{1\;{\text{J}}}}{{{\text{1}}\;{\text{kg}}\cdot {{\text{m}}^2}{\text{/}}{{\text{s}}^2}}}} \right)\\ &=2.6937\; \times {10^{ - 11}}\;{\text{J}}\\\end{aligned}$