Question

How much work is needed to assemble an atomic nucleus containing three protons (such as Li) if we model it as an equilateral triangle of side 2.00×10−15m2.00×10−15m with a proton at each vertex? Assume the protons started from very far away.

Answer 1

Answer:

2.1576 MeV

Explanation:

r = Distance = $2\times 10^{-15}\ m$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge of electron = $1.6\times 10^{-19}\ C$

The resulting energy of the system is

$U=\dfrac{k3q^2}{r}\\\Rightarrow U=\dfrac{8.99\times 10^9\times 3(1.6\times 10^{-19})^2}{2\times 10^{-15}}\\\Rightarrow U=3.45216\times 10^{-13}\ J$

Converting to MeV

$\dfrac{3.45216\times 10^{-13}}{1.6\times 10^{-13}}=2.1576\ MeV$

The work needed to assemble an atomic nucleus is 2.1576 MeV

Answer 2

All things in the universe are made up of elements. There are different types of elements in our surroundings. These elements are as follows:-

• Carbon
• Hydrogen
• Oxygen and etc

According to the question, the formula used in the question is $U= \frac{k3q^{2} }{r}$

As the data is given in the question the solution is:-

$U =\frac{8.99*10^{9} * )3(1.6 *10^{-19})^{2} }{2* 10^{-15} }$

After solving the question, the value of U we get is $U =3.45216 \ X \ 10^{-13} J$

The joule must be converted into Mev.Hence, the solution is -$\frac{3.45216 \ X \ 10^{-13} J}{1.6*10^{-13} } = 2.1576 Mev$

For more solutions, refer to this link:-

https://brainly.com/question/23379286