The Bohr model of the hydrogen atom pictures the electron as a tinyparticle movi

Question

The Bohr model of the hydrogen atom pictures the electron as a tinyparticle moving in a circular orbit about a stationary proton. Inthe lowest-energy orbit the distance from the proton to theelectron is 5.29 x 10^-11m, and the linear speed of the electron is2.18 x 10^6 m/s.

a)What is the angular speed of the electron (rad/s)?
b)How many orbits about the proton does it make each second(rev/s)?
c) What is the electron's centripetal acceleration (m/s^2)?

I really don't understand how to do this. I know that the eq. forspeed is delta theta / delta time but since those are not givenvariables i'm not sure what to do for any of the parts.

Explanation / Answer

a) The angular pseed of the electron is                 = v/r                    = ( 2.18 x 10^6 m/s)/(5.29 x 10^-11m)                   =---- rad/s b)   The number of orbits about the protondoes it make each second is                N =  ( 2.18 x 10^6 m/s)/(5.29 x 10^-11m) (1rev/2)                     =----- rev/s c) the electron's centripetal acceleration is                a = v2/r                    = (2.18 x 10^6 m/s)2/(5.29 x 10^-11m)                   =----- m/s2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---