Answer:
The orbital speed of the electron is 2.296 x 10⁶ m/s
Explanation:
Given;
radius of the circular orbit, r = 5.041 × 10⁻¹¹ m
In the Bohr’s model of the hydrogen atom, the velocity of the electron is given as;
[tex]V = \frac{nh}{2\pi mr}[/tex]
where;
h is Planck's constant
m is mass of electron
r is the radius of the circular orbit
n is the energy level of hydrogen in ground state
Substitute in these values and solve for V
[tex]V = \frac{nh}{2\pi mr} = \frac{1*6.626*10^{-34}}{2\pi *9.11 *10^{-31}*5.041*10^{-11}}\\\\V = 2.296*10^6 \ m/s[/tex]
Therefore, the orbital speed of the electron is 2.296 x 10⁶ m/s
