In the Bohr model of the hydrogen atom, an electron (mass = 9.10 10-31 kg) orbits a proton at a distance of 4.76x10-10 m. The proton pulls on the electron with an electric force of 1.02x10-9 N. How many revolutions per second does the electron make?

Question

In the Bohr model of the hydrogen atom, an electron (mass = 9.10  10-31 kg) orbits a proton at a distance of 4.76x10-10 m. The proton pulls on the electron with an electric force of 1.02x10-9 N. How many revolutions per second does the electron make?

Sebastian Acosta · Accepted Answer

We know that the force of attraction #F#of  proton on electron in the hydrogen atom provides  the required centripetal force required for the motion of electron in circular orbit. If #n# be the number of revolutions per sec made by the electron around the nucleus then centripetal force is given by #F=momega^2r=m4pi^2n^2r#  where radius of the orbit #" "r=4.76xx10^-10m# mass of an electron #" "m=9.1xx10^-31kg# #F=1.02xx10^-19N# So #m4pi^2n^2r=F#  #=>n^2=F/(4pi^2xxmxxr)# #=>n=sqrt(F/(4pi^2xxmxxr))# #=sqrt((1.02xx10^-19N)/(4pi^2xx9.1xx10^-31kgxx4.76xx10^-10m))# #=3.74xx10^8Hz#

Isabella Alvarez · Answer

undefined The frequency of the electron's revolutions can be calculated using the formula:

$$ f = \frac{v}{2\pi r} $$

Where $ f $ is the frequency, $ v $ is the velocity of the electron, and $ r $ is the radius of its orbit.

The velocity of the electron can be determined using the centripetal force equation:

$$ F = \frac{mv^2}{r} $$

Solving for $ v $:

$$ v = \sqrt{\frac{Fr}{m}} $$

Substituting the given values:

$$ v = \sqrt{\frac{(1.02 \times 10^{-9} \, N) \times (4.76 \times 10^{-10} \, m)}{9.10 \times 10^{-31} \, kg}} $$

$$ v \approx 2.19 \times 10^6 \, m/s $$

Now, substituting $ v $ into the frequency formula:

$$ f = \frac{2.19 \times 10^6 \, m/s}{2 \pi \times 4.76 \times 10^{-10} \, m} $$

$$ f \approx 3.67 \times 10^{15} \, Hz $$

So, the electron makes approximately $ 3.67 \times 10^{15} $ revolutions per second.