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Chapter 11: Problem 1557

If electron in ground state of \(\mathrm{H}\) -atom is assumed in rest then dipole moment of electron proton system of \(\mathrm{H}\) -atom is $\ldots \ldots\( Orbit radius of \)\mathrm{H}\( atom in ground state is \)0.56 \AA$. (A) \(0.253 \times 10^{-29} \mathrm{~m}\) (B) \(0.848 \times 10^{-29} \mathrm{~m}\) (C) \(0.305 \times 10^{-29} \mathrm{~m}\) (D) \(1.205 \times 10^{-28} \mathrm{~m}\)

Short Answer

Expert verified
The dipole moment of the electron-proton system in the hydrogen atom is approximately \(0.848 \times 10^{-29} \mathrm{~m}\) (Option B).

Step by step solution

01

Identify given information

We are given the orbit radius of a hydrogen atom in the ground state: Orbit radius = 0.56 Å = 0.56 × 10^{-10} meters
02

Find the charge of an electron and proton

The charge of an electron and proton is equal in magnitude but opposite in sign. For our calculation, we use the magnitude of the charge: Charge of electron (or proton) = 1.6 × 10^{-19} Coulombs
03

Calculate the dipole moment

Now we can use the formula for dipole moment: dipole moment = charge × distance Dipole moment = (1.6 × 10^{-19} C) × (0.56 × 10^{-10} m)
04

Evaluate the answer

Evaluate the numerical expression to find the dipole moment of the electron-proton system in the hydrogen atom. Dipole moment = \( (1.6 \times 10^{-19})(0.56 \times 10^{-10}) \) = 0.896 × 10^{-29} Cm Now we compare this result to the options given in the exercise: (A) 0.253 × 10^{-29} m (B) 0.848 × 10^{-29} m (C) 0.305 × 10^{-29} m (D) 1.205 × 10^{-28} m The closest option to our calculated value is (B) 0.848 × 10^{-29} m, which would be the dipole moment of the electron-proton system in the hydrogen atom.

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