Question

Angular speed of an electron in the ground state of hydrogen atom is 4 × 10¹⁶ rad/s. What is its angular speed in 4^th orbit?

पर्याय

• 4.25 × 10¹⁴ rad/s

• 6.25 × 10¹⁴ rad/s

• 6.25 × 10¹⁶ rad/s

• 0.25 × 10¹⁶ rad/s

Answer 1

6.25 × 10¹⁴ rad/s

Explanation:

`omega prop 1/"n"^3`

`therefore omega_2/omega_1 = (1/4)^3 = 1/64`

`therefore omega_2 = omega_1/64 = (4 xx 10^16)/64 = 6.25 xx 10^16`rad/s