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

Numerical

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

#### Solution

Data: z = 1, m = 9.1 x 10-31 kg, e = 1.6 x 10-19 C, ε0 = 8.85 x 10-12 C2/N·m2, h = 6.63 x 10-34 J.s, n = 2, t = 10-8 s

The periodic time of the electron in a hydrogen atom,

T = (4ε_0^2"h"^3"n"^3)/(pi"me"^4)

= ((4)(8.85 xx 10^-12)^2(6.63 xx 10^-34)^3(8))/((3.142)(9.1 xx 10^-31)(1.6 xx 10^-19)^4)

= ((4)(8.85)^2(6.63)^3(8))/((3.142)(9.1)(1.6)^4) xx 10^-19s

= 3.898 × 10-16 s

Let N be the number of revolutions made by the electron in time t. Then, t =NT.

∴ N = "t"/"T" = 10^-8/(3.898 xx 10^-16) = 2.565 xx 10^7

Concept: Mass-energy and Nuclear Binding Energy - Nuclear Binding Energy
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 15 Structure of Atoms and Nuclei
Exercises | Q 11 | Page 342