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?

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#### Solution

**Data:** z = 1, m = 9.1 x 10^{-31} kg, e = 1.6 x 10^{-19} C, ε_{0} = 8.85 x 10^{-12} C^{2}/N·m^{2}, 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

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