Question

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm. (a) Find the magnitude of the electrostatic force of attraction, Fe between the electron and the proton. (b) Find the magnitude of the gravitational force of attraction Fg , between the electron and the proton, and find the ratio, Fe /Fg . me = 9.11 x 10-31kg, e = 1.602 x 10-19C mp = 1.67 x 10-27kg k = 9 x 109 Nm2 /C2 G = 6.67 x 10-11 Nm2 /kg2

Answer 1

(a):
`F_e = 8.202* 10^(-8)\ \rm N.`

(b):
`F_g = 3.6125* 10^(-47)\ \rm N.`

(c):
`(F_e)/(F_g)=2.27* 10^(39).`

Step-by-step explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053
`* 10^(-9)` m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges
`q_1` and
`q_2` respectively is given by

`F_e = (k|q_1||q_2|)/(r^2)`

where,

• 
  `k` = Coulomb's constant =
  `9* 10^9\ \rm Nm^2/C^2.`
• 
  `r` = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom,
`q_1 = +1.6* 10^(-19)\ C.`

The charge on the electron,
`q_2 = -1.6* 10^(-19)\ C.`

These two are separated by the distance,
`r = 0.053* 10^(-9)\ m.`

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

`F_e = ((9* 10^9)* |+1.6* 10^(-19)|* |-1.6* 10^(-19)|)/((0.053* 10^(-9))^2)=8.202* 10^(-8)\ \rm N.`

Part (b):

The gravitational force of attraction between two objects of masses
`m_1` and
`m_1` respectively is given by

`F_g = (Gm_1m_2)/(r^2).`

where,

• 
  `G` = Universal Gravitational constant =
  `6.67* 10^(-11)\ \rm Nm^2/kg^2.`
• 
  `r` = distance of separation between the masses.

For the given system,

The mass of proton,
`m_1 = 1.67* 10^(-27)\ kg.`

The mass of the electron,
`m_2 = 9.11* 10^(-31)\ kg.`

Distance between the two,
`r = 0.053* 10^(-9)\ m.`

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

`F_g = ((6.67* 10^(-11))* (1.67* 10^(-27))* (9.11* 10^(-31)))/((0.053* 10^(-9))^2)=3.6125* 10^(-47)\ \rm N.`

The ratio
`(F_e)/(F_g)`:

`(F_e)/(F_g)=(8.202* 10^(-8))/(3.6125* 10^(-47))=2.27* 10^(39).`