A hydrogen atom contains one proton and one electron. It may be assumed that the electron revolves in a circle of radius 0.53 angstrom `(1 angstrom = 10^(-10)` m and is abbreviated as A) with the proton at the center. The hydrogen atom is said to be in the ground state in this case. Find the speed of the electron in the ground state of hydrogen atom.

Explanation

`F_e` from previous problem `No.18`
` =8.2xx10^-8N, V=? `
Now, `M_e = 9.12xx10^-31kg,`
` R=0.53xx10^-10m `
Now, `F_e =(M_eV^2 )/R `
` So, V^2=0.4775xx10^13 `
` =4.775xx10^12m^2//s^2 `
` V=2.18xx10^6 m//s`