Question

A hydrogen atom is at the earth’s surface. The electron and proton in the atom are separated by a dis- tance of 5.29 3 10211 m. What is the ratio of the magnitude of the electrical force exerted by the proton on the electron to the weight of the electron?

Answer 1

`(F_e)/(F_(g)) = 9.87 * 10^(21)`

Step-by-step explanation:

As we know that the charge on electron and proton is same as the charge of an electron

so the electrostatic force between the electron and proton is given as

`F = (kq_1q_2)/(r^2)`

so we will have

`F = ((9* 10^9)(1.6 * 10^(-19))^2)/((5.11 * 10^(-11))^2)`

`F = 8.82 * 10^(-8) N`

Now the weight of an electron is given as

`W = mg`

`W = (9.11 * 10^(-31))(9.81)`

`W = 8.94 * 10^(-30) N`

now the ratio of electrical force and weight of the electron is given as

`(F_e)/(F_g) = (8.82 * 10^(-8))/(8.94 * 10^(-30))`

`(F_e)/(F_(g)) = 9.87 * 10^(21)`