Advertisement Remove all ads

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?

Advertisement Remove all ads

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^-19`s

= 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
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

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

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×