Question

An electron in hydrogen atom-stays in its second orbit for 10⁻⁸ s. How many revolutions will it make around the nucleus in that time?

[Velocity of electron in second orbit = 1.07 × 10⁶ m/s, radius of electron in second orbit = 2.14 × 10⁻¹⁰ m]

Answer 1

V₂ = 1.07 × 10⁶ m/s, r₂ = 2.14 × 10⁻¹⁰ m, t = 10⁻⁸ s, π = 3.142

Number of revolutions made in time t = ft

= `(V_2t)/(2πr_2)`

= `(1.07 xx 10^6 xx 10^-8)/(2 xx 3.142 xx 2.14 xx 10^-10)`

= `(107 xx 10^6)/(3.142 xx 4.28)`

= 7.957 × 10⁶