If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion HH2+. In the ground state of an HH2+, the two protons are separated - Physics

Numerical

If one of the two electrons of a Hmolecule is removed, we get a hydrogen molecular ion "H"_2^+. In the ground state of an "H"_2^+, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of zero potential energy.

Solution

The system of two protons and one electron is represented in the given figure.

Charge on proton 1, q1 = 1.6 × 10−19 C

Charge on proton 2, q2 = 1.6 × 10−19 C

Charge on electron, q3 = −1.6 × 10−19 C

Distance between protons 1 and 2, d1 = 1.5 × 10−10 m

Distance between proton 1 and electron, d2 = 1 × 10−10 m

Distance between proton 2 and electron, d3 = 1 × 10−10 m

The potential energy at infinity is zero.

Potential energy of the system,

"V" = ("q"_1"q"_2)/(4piin_0"d"_1) + ("q"_2"q"_3)/(4piin_0"d"_3) + ("q"_3"q"_1)/(4piin_0"d"_2)

Substituting 1/(4piin_0) = 9 xx 10^9  "N m"^2   "C"^-2, we obtain

"V" = (9 xx 10^9 xx 10^-19 xx 10^-19)/(10^-10) [-(16)^2 + (1.6)^2/(1.5) + -(1.6)^2]

= -30.7 xx 10^-19  "J"

= −19.2 eV

Therefore, the potential energy of the system is −19.2 eV.

Concept: Potential Energy of a System of Charges
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NCERT Physics Part 1 and 2 Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.19 | Page 88
NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 19 | Page 90
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