A hydrogen atom contains one proton and one electron- It may be assumed that the electron revolves in a circle of radius 0-53 angstrom 1angstrom - 10-10 m and is abbreviated as A with the proton at the centre- Teh hydrogen atom is said to be in the ground state in this case- Find th magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state-

R = 0.53∘A = 0.53 × 10^ 10m F = 9 × 10^9 × (1.6 × 10^ 19)^2/(0.53 × 10^ 10)^2 = 8.2 × 10^ 8N ...

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Standard XII

Physics

Coulomb's Law - Mathematical Form

A hydrogen at...

Question

A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0.53 angstrom $(1 a n g s t r o m = 10^{- 10}$ m and is abbreviated as $\circ A$ with the proton at the centre. Teh hydrogen atom is said to be in the ground state in this case. Find th magnitude of the electric force between the proton and the electron of a hydrogen atom in its ground state.

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Solution

$R = 0.53 \circ A = 0.53 \times 10^{- 10} m F = \frac{9 \times 10^{9} \times (1.6 \times 10^{- 19})^{2}}{(0.53 \times 10^{- 10})^{2}} = 8.2 \times 10^{- 8} N$

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R=0.53∘A=0.53×10−10mF=9×109×(1.6×10−19)2(0.53×10−10)2=8.2×10−8N

$R = 0.53 \circ A = 0.53 \times 10^{- 10} m F = \frac{9 \times 10^{9} \times (1.6 \times 10^{- 19})^{2}}{(0.53 \times 10^{- 10})^{2}} = 8.2 \times 10^{- 8} N$