Question

in the bhor model of the hydrogen atom the electron is assumed to orbit the proton in a circle at an average distance of 5.3x10^-11 m. the centripetal force keeping the electron in orbit is due to the Coulomb force law. e= 1.6x10^-19C, electron = 9.11 x 10^-31

1. what is the electrons velocity?

2. how long does it take the electron to make one orbit around the proton?

3. the current produced by the electron orbiting the proton is equal to the charge of the electron divided by the orbital period. determine the current produced by the electron.

4. what is the magnitude of the magnetic field produced by the electron at the center of its orbit?

Answer #1

1)

Equating Coulomb force with centripetal force,

ke^{2}/r^{2} = mv^{2}/r

Where k = 1/4o,
e the charge of electron, r is the orbit radius, m is the mass of
electron and v be the velocity.

mv^{2} = ke^{2}/r

v^{2} = ke^{2}/mr

v = SQRT[ke^{2}/mr]

= SQRT{[(8.99 x 10^{9}) x (1.6 x
10^{-19})^{2}] / [(9.11 x 10^{-31}) x (5.3
x 10^{-11})]}

= 2.18 x 10^{6} m/s

2)

Distance traveled = 2r

= 2
x (5.3 x 10^{-11}) = 3.33 x 10^{-10} m

Velocity, v = 2.18 x 10^{6} m/s

Time taken, T = Distance/velocity

= (3.33 x 10^{-10}) / (2.18 x 10^{6})

= 1.53 x 10^{-16} s

3)

Current, I = e/T

= (1.6 x 10^{-19}) / (1.53 x 10^{-16})

= 1.05 x 10^{-3} A

4)

Magnetic field, B = _{
}oI/2r

= [(4
x 10^{-7}) x (1.05 x 10^{-3})] / [2 x (5.3 x
10^{-11})]

= 12.4 T

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