2.19 If one of the two electrons of a <math><msub><mrow><mrow><mi mathvariant="bold-italic">H</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> molecule is removed, we get a hydrogen molecular ion <math><msubsup><mrow><mrow><mi mathvariant="bold-italic">H</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mo>+</mo></mrow></mrow></msubsup></math> . In the ground state of an <math><msubsup><mrow><mrow><mi mathvariant="bold-italic">H</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mo>+</mo></mrow></mrow></msubsup></math> , 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 the zero of potential energy.

2.19Let the charge in Proton 1 be q1 = 1.6 ×10-19 C, in Proton 2, q2 = 1.6 ×10-19 C and the charge in the electron, q3 = - 1.6 ×10-19 CDistance between Proton 1 & Proton 2, d1 = 1.5 Å = 1.5 ×10-10 mDistance between Proton 1 and Electron, d2 = 1.0 Å = 1.0 ×10-10 mDistance between Proton 2 and Electron, d3 = 1.0 Å = 1.0 ×10-10 mThe potential energy of the system,V = q1q24πε0d1 + q2q34πε0d3 + q3q14πε0d2 ,Where ε0=Permittivityoffreespace = 8.854 ×10-12 C2N-1 m-2V = 14πε0 ( q1q2d1 + q2q3d2 + q3q1d2 )= 14π×8.854×10-12× ( 1.6×10-19×1.6×10-191.5×10-10 + 1.6×10-19×-1.6×10-191.0×10-10 + -1.6×10-19×1.6×10-191.0×10-10 )= 8.988 ×109×(1.707×10-28 -2.56×10-28-2.56×10-28)= -3.067×10-18 J= -3.067×10-18×11.6×10-19 eV= - 19.17 eV

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Physics Ncert Solutions Class 12th Physics Class 12th Electrostatic Potential and Capacitance

2.19 If one of the two electrons of a $H_{2}$ molecule 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 the zero of potential energy.

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alok kumar singh | Contributor-Level 10

3 months ago

2.19

Let the charge in Proton 1 be $q_{1}$ = 1.6 $\times 10^{- 19}$ C, in Proton 2, $q_{2}$ = 1.6 $\times 10^{- 19}$ C and the charge in the electron, $q_{3}$ = $-$ 1.6 $\times 10^{- 19}$ C

Distance between Proton 1 & Proton 2, $d_{1}$ = 1.5 Å = 1.5 $\times 10^{- 10}$ m

Distance between Proton 1 and Electron, $d_{2}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

Distance between Proton 2 and Electron, $d_{3}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

The potential energy of the system,

V = $\frac{q_{1} q_{2}}{4 π ε_{0} d_{1}}$ + $\frac{q_{2} q_{3}}{4 π ε_{0} d_{3}}$ + $\frac{q_{3} q_{1}}{4 π ε_{0} d_{2}}$ ,

Where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

V = $\frac{1}{4 π ε_{0}}$ ( $\frac{q_{1} q_{2}}{d_{1}}$ + $\frac{q_{2} q_{3}}{d_{2}}$ + $\frac{q_{3} q_{1}}{d_{2}}$ )

= $\frac{1}{4 π \times 8.854 \times 10^{- 12}} \times$ ( $\frac{1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{1.5 \times 10^{- 10}}$ + $\frac{1.6 \times 10^{- 19} \times - 1.6 \times 10^{- 19}}{1.0 \times 10^{- 10}}$ +&n

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2.19

Let the charge in Proton 1 be $q_{1}$ = 1.6 $\times 10^{- 19}$ C, in Proton 2, $q_{2}$ = 1.6 $\times 10^{- 19}$ C and the charge in the electron, $q_{3}$ = $-$ 1.6 $\times 10^{- 19}$ C

Distance between Proton 1 & Proton 2, $d_{1}$ = 1.5 Å = 1.5 $\times 10^{- 10}$ m

Distance between Proton 1 and Electron, $d_{2}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

Distance between Proton 2 and Electron, $d_{3}$ = 1.0 Å = 1.0 $\times 10^{- 10}$ m

The potential energy of the system,

V = $\frac{q_{1} q_{2}}{4 π ε_{0} d_{1}}$ + $\frac{q_{2} q_{3}}{4 π ε_{0} d_{3}}$ + $\frac{q_{3} q_{1}}{4 π ε_{0} d_{2}}$ ,

Where $ε_{0} = P e r m i t t i v i t y o f f r e e s p a c e$ = 8.854 $\times 10^{- 12}$ $C^{2} N^{- 1}$ $m^{- 2}$

V = $\frac{1}{4 π ε_{0}}$ ( $\frac{q_{1} q_{2}}{d_{1}}$ + $\frac{q_{2} q_{3}}{d_{2}}$ + $\frac{q_{3} q_{1}}{d_{2}}$ )

= 8.988 $\times 10^{9} \times (1.707 \times 10^{- 28}$ $- 2.56 \times 10^{- 28} - 2.56 \times 10^{- 28})$

= $- 3.067 \times 10^{- 18}$ J

= $- 3.067 \times 10^{- 18} \times \frac{1}{1.6 \times 10^{- 19}}$ eV

= - 19.17 eV

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