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Question: In an atom, two protons are bounded by a distance of \( 2.8 \times {10^{ - 18}}m \) , and an electro...

In an atom, two protons are bounded by a distance of 2.8×10−18m2.8 \times {10^{ - 18}}m , and an electron is at a distance of 1.4×10−18m1.4 \times {10^{ - 18}}m from each proton. Calculate the potential energy of the system. (Given charge of an electron −1.6×10−19C- 1.6 \times {10^{ - 19}}C and charge of proton +1.6×10−19C+ 1.6 \times {10^{ - 19}}C ).

Explanation

Solution

To solve this question, we have to divide the given system into three systems each containing a pair of two charges. Then we have to calculate the potential energy for these systems and finally, we have to add the potential energies to get the total potential energy of the system.
Formula used: The formula used to solve this question is
⇒U=kq1q2r\Rightarrow U = \dfrac{{k{q_1}{q_2}}}{r} , where UU is the potential energy of a system of two charges q1{q_1} and q2{q_2} , separated by a distance rr

Complete step by step answer:
We know that the potential energy for a pair of two charges is given by
⇒U=kq1q2r\Rightarrow U = \dfrac{{k{q_1}{q_2}}}{r}
As the system contains one electron and two protons, we can divide the system into three pairs of charges. One pair contains two protons, and the other two pairs contain an electron and a proton.
For the first system
For this case we have, q1=q2=1.6×10−19C{q_1} = {q_2} = 1.6 \times {10^{ - 19}}C and r=2.8×10−18mr = 2.8 \times {10^{ - 18}}m
So, U1=k(1.6×10−19)22.8×10−18{U_1} = \dfrac{{k{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{2.8 \times {{10}^{ - 18}}}}
On solving we get
⇒U1=k(9.14×10−21)\Rightarrow {U_1} = k\left( {9.14 \times {{10}^{ - 21}}} \right)
For the second system
For this case we have, q1=−1.6×10−19C{q_1} = - 1.6 \times {10^{ - 19}}C q2=1.6×10−19C{q_2} = 1.6 \times {10^{ - 19}}C and r=1.4×10−18mr = 1.4 \times {10^{ - 18}}m
So, U2=−k(1.6×10−19)21.4×10−18{U_2} = \dfrac{{ - k{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{1.4 \times {{10}^{ - 18}}}}
On solving we get
⇒U2=−k(18.28×10−21)\Rightarrow {U_2} = - k\left( {18.28 \times {{10}^{ - 21}}} \right)
As the third system is identical to the second system, so U3=U2=−k(18.28×10−21){U_3} = {U_2} = - k\left( {18.28 \times {{10}^{ - 21}}} \right)
Now, the total potential energy of the system
⇒U=U1+U2+U3\Rightarrow U = {U_1} + {U_2} + {U_3}
⇒U=k(9.14×10−21)−k(18.28×10−21)−k(18.28×10−21)\Rightarrow U = k\left( {9.14 \times {{10}^{ - 21}}} \right) - k\left( {18.28 \times {{10}^{ - 21}}} \right) - k\left( {18.28 \times {{10}^{ - 21}}} \right)
On simplifying, we get
⇒U=−k(27.42×10−21)\Rightarrow U = - k\left( {27.42 \times {{10}^{ - 21}}} \right)
As we know that k=9×109k = 9 \times {10^9}
⇒U=−9×109×27.42×10−21\Rightarrow U = - 9 \times {10^9} \times 27.42 \times {10^{ - 21}}
⇒U=−2.47×10−10J\Rightarrow U = - 2.47 \times {10^{ - 10}}J
Hence, the potential energy of the system is equal to −2.47×10−10J- 2.47 \times {10^{ - 10}}J .

Note:
Do not ignore the negative sign on the potential energy. As we all know that the potential energy is a scalar quantity. So the negative sign does not represent any direction, but it is a part of its magnitude. Ignoring the negative sign will lead to an incorrect answer.