# Find the speed of the electron in the ground state of a hydrogen atom. The description of ground state is given in the previous problem.

Kevin Nash
10 years ago
Sol Fe from previous problem No. 18 = 8.2 × 10^–8 N Ve = ? Now, M base e = 9.12 × 10^–31 kg r = 0.53 × 10^–10 m Now, Fe = M base ev^2/r ⇒ v^2 = Fe * r/m base e = 8.2 * 10^-8 * 0.53 * 10^-10/9.1 * 10^-31 = 0.4775 * 10^13 = 4.775 * 10^12 m^2/s^2 ⇒ v = 2.18 * 10^6 m/s
Deepak Patra
10 years ago
Sol. Fe from previous problem No. 18 = 8.2 × 10^–8 N Ve = ? Now, M base e = 9.12 × 10^–31 kg r = 0.53 × 10^–10 m Now, Fe = M base ev^2/r ⇒ v^2 = Fe * r/m base e = 8.2 * 10^-8 * 0.53 * 10^-10/9.1 * 10^-31 = 0.4775 * 10^13 = 4.775 * 10^12 m^2/s^2 ⇒ v = 2.18 * 10^6 m/s
Rishi Sharma
4 years ago
Dear Student,

Fe from previous problem No. 18 = 8.2 × 10^–8 N Ve = ?
Now, M base e = 9.12 × 10^–31 kg
r = 0.53 × 10^–10 m
Now, Fe = M base ev^2/r
⇒ v^2 = Fe * r/m base e = 8.2 * 10^-8 * 0.53 * 10^-10/9.1 * 10^-31
= 0.4775 * 10^13
= 4.775 * 10^12 m^2/s^2
⇒ v = 2.18 * 10^6 m/s

Thanks and Regards