In atomic physics, the Bohr model depicts the hydrogen atom as a positively char

Question

In atomic physics, the Bohr model depicts the hydrogen atom as a positively charged proton surrounded by an electron that travels in a circular orbit around the proton. If the electron orbits hte proton at a distance of 5.3x10^-11m and completes one revolution in 1.5x10^-16 seconds, what is the force of attraction between the proton and electron? The mass of the electron is 9.1x10^-31 --------- I found our the equation is F=(4pi^2mr)/t^2 .... I just don't know the values to plug it in to ... The answer is suppose to be 8.5x10^-8N

Explanation / Answer

Radius of the circular orbit (r) =5.3x10^-11 m Distance covered in one rotation = 2*pi*r =33.284 x 10^-11 time taken to travel =1.5 x 10^-16 s therefore velocity of electron (v) = 33.284 x 10^-11/1.5x10^-16 =22.19 x 10^5 m/s Force of attraction causes circular motion, therefore force force = m.v^2/r force =9.1x10^-31x(22.19x10^5)^2/(5.3x10^-11) force=845.43 x 10^-10 force=8.45 X 10^ -8 answer ::)

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