Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# The Average Separation Between the Proton and the Electron in a Hydrogen Atom in Ground State is 5.3 × 10−11 M. (A) Calculate the Coulomb Force Between Them at this Separation. - Physics

Sum

The average separation between the proton and the electron in a hydrogen atom in ground state is 5.3 × 10−11 m. (a) Calculate the Coulomb force between them at this separation. (b) When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state?

#### Solution

Average separation between the proton and the electron of a Hydrogen atom in ground state, r = 5.3 × 10−11

(a) Coulomb force when the proton and the electron in a hydrogen atom in ground state

$F = 9 \times {10}^9 \times \frac{q_1 q_2}{r_2}$

$= \frac{9 \times {10}^9 \times \left( 1 . 6 \times {10}^{- 19} \right)^2}{\left( 5 . 3 \times {10}^{- 11} \right)^2} = 8 . 2 \times {10}^{- 8} \text{N}$

(b) Coulomb force when the average distance between the proton and the electron becomes 4 times that of its ground state

$\text {Coulomb force, F } = \frac{1}{4\pi \in_0} = \frac{q_1 q_2}{\left( 4r \right)^2}$

$= \frac{9 \times {10}^9 \times \left( 1 . 6 \times {10}^{- 19} \right)^2}{16 \times \left( 5 . 3 \right)^2 \times {10}^{- 22}}$

$= \frac{9 \times \left( 1 . 6 \right)^2}{10 \times \left( 5 . 3 \right)^2} \times {10}^{- 7}$

$= 0 . 0512 \times {10}^{- 7}$

$= 5 . 1 \times {10}^{- 9} \text{ N }$

Concept: Work Done by a Constant Force and a Variable Force
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 4 The Forces
Exercise | Q 11 | Page 63
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