Question

The average separation between the proton and the electron in a hydrogen atom in ground state is 5.3 x 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?
Concept of Physics - 1 , HC VERMA , Chapter "The Force"

Answer 1

Hello Dear.

Given in the question :-

The average separation b/w the proton & the electron in a hydrogen atom in ground state is = 5.3 x 10⁻¹¹ .
∴ r = 5.3 x 10⁻¹¹ .

(a) For calculating the coulomb force b/w them at this separation :- 

Coulomb force, F₁ = 1/4πε₀ $\frac{q_{1} q _{2} }{r^2}$ 
F₁ = 9 x 10⁹ x q₁q₂/r²
F₁= 9x 10⁻⁹ x (1.6 x 10⁻¹⁹)²/(5.3 x 10⁻¹¹)²    [∵q₁=q₂ = 1.6 x 10⁻¹⁹]
F₁ =  9 x 10⁻⁹ x {2.56 x 10⁻³⁸ / 2.81 x 10⁻²²}

F₁ =  8.2 x 10⁻⁸ N.

The coulomb force b/w separation is 8.2 x 10⁻⁸ N.


(b) when atom is in first exited state , the coulomb force in that state :-

Here given that, The average distance b/w proton and electron is 4 times higher then it's ground state.

Now according to the given condition.

F₂ = F₁ / 16
F₂ = 8.2 x10⁻⁸ / 16
F₂ = 5.125 x 10⁻⁹ N. 

Hence the coulomb force in this state is equal to 5.125 x 10⁻⁹ N.



Hope it Helps. :-)

Answer 2

Answer:

(a)8.2 × 10^-8 N

(b)5.1 × 10^-9 N

Explanation:

Explanation in the above attachment.