Advertisements
Advertisements
Question
Angular speed of an electron in the ground state of hydrogen atom is 4 × 1016 rad/s. What is its angular speed in 4th orbit?
Options
4.25 × 1014 rad/s
6.25 × 1014 rad/s
6.25 × 1016 rad/s
0.25 × 1016 rad/s
MCQ
Advertisements
Solution
6.25 × 1014 rad/s
Explanation:
`omega prop 1/"n"^3`
∴ `omega_2/omega_1 = (1/4)^3 = 1/64`
∴ `omega_2 = omega_1/64`
= `(4 xx 10^16)/64`
= `6.25 xx 10^14`rad/s
shaalaa.com
Bohr’s Atomic Model
Is there an error in this question or solution?
