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?

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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?

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Answer 1 · 2016-12-10T10:11:03

We know that the force of attraction $F$ $F$ of proton on electron in the hydrogen atom provides the required centripetal force required for the motion of electron in circular orbit.

If $n$ $n$ be the number of revolutions per sec made by the electron around the nucleus then centripetal force is given by

$F = m ω^{2} r = m 4 π^{2} n^{2} r$ $F=momega^2r=m4pi^2n^2r$

where radius of the orbit $r = 4.76 \times 10^{- 10} m$ $" "r=4.76xx10^-10m$

mass of an electron $m = 9.1 \times 10^{- 31} k g$ $" "m=9.1xx10^-31kg$

$F = 1.02 \times 10^{- 19} N$ $F=1.02xx10^-19N$

So

$m 4 π^{2} n^{2} r = F$ $m4pi^2n^2r=F$

$\Rightarrow n^{2} = \frac{F}{4 π^{2} \times m \times r}$ $=>n^2=F/(4pi^2xxmxxr)$

$\Rightarrow n = \sqrt{\frac{F}{4 π^{2} \times m \times r}}$ $=>n=sqrt(F/(4pi^2xxmxxr))$

$= \sqrt{\frac{1.02 \times 10^{- 19} N}{4 π^{2} \times 9.1 \times 10^{- 31} k g \times 4.76 \times 10^{- 10} m}}$ $=sqrt((1.02xx10^-19N)/(4pi^2xx9.1xx10^-31kgxx4.76xx10^-10m))$

$= 3.74 \times 10^{8} H z$ $=3.74xx10^8Hz$

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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?

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