In a hydrogen atom,the distance between the e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The electrical force of attraction between the electron and proton is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.Here,$k = 9 	imes 10^

Vedclass · Accepted Answer

$3.7 	imes 10^{-7} \ N$

Vedclass · Answer

(B) The electrical force of attraction between the electron and proton is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.Here,$k = 9 	imes 10^9 \ N \cdot m^2/C^2$,$q_1 = q_2 = 1.6 	imes 10^{-19} \ C$,and $r = 2.5 	imes 10^{-11} \ m$.Substituting the values:$F = \frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19}) 	imes (1.6 	imes 10^{-19})}{(2.5 	imes 10^{-11})^2}$$F = \frac{9 	imes 2.56 	imes 10^{9 - 19 - 19}}{6.25 	imes 10^{-22}}$$F = \frac{23.04 	imes 10^{-29}}{6.25 	imes 10^{-22}}$$F = 3.6864 	imes 10^{-7} \ N \approx 3.7 	imes 10^{-7} \ N$.

In a hydrogen atom,the distance between the electron and proton is $2.5 \times 10^{-11} \ m$. The electrical force of attraction between them will be

Similar Questions

The dimensional formula of $k$ (Coulomb's constant) is . . . . . . .
(Take $I$ as the dimension of current.)

$12$ positive charges of magnitude $q$ are placed on a circle of radius $R$ in a manner that they are equally spaced. $A$ charge $Q$ is placed at the centre. If one of the charges $q$ is removed,then the force on $Q$ is

Two charges $Q$ are placed at opposite corners of a square. Two charges $q$ are placed at the other two corners. If the net electric force on $Q$ is zero,then $Q/q$ is equal to:

Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is

Vedclass Test Series

Exam Paper Generator

Online Exam Module