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Average Lifetime of a Hydrogen Atom Excited to N = 2 State is 10−8 S. Find the Number of Revolutions Made by the Electron on the Average before It Jumps to the Ground State.

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Question

Average lifetime of a hydrogen atom excited to n = 2 state is 10−8 s. Find the number of revolutions made by the electron on the average before it jumps to the ground state.

Sum
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Solution

Frequency of electron (f) is given by

`f = (me^4)/(4∈_0^2n^3h^3)`

Time period is given by

` T = 1/f`

`T = (4∈_0^2n^3h^3)/me^`

Here,

h = Planck's constant

m = Mass of the electron

e = Charge on the electron

`∈_0`= Permittivity of free space

`∴ T =(4xx(8.85xx10^-12)xx(2)^3xx(6.63xx10^-34)^3)/((9.10xx10^-31)xx(1.6xx106^-16)^4 `

`= 12247.735xx10^-19  S`

Average life time of hydrogen, `t = 10^-3 S`
Number of revolutions is given by

`N = t/T`

`rArr N=(10^-8)/(12247.735xx10^-19)`

N = 8.2 × 10 revolution

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Chapter 43: Bohr’s Model and Physics of Atom - Exercises [Page 385]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 43 Bohr’s Model and Physics of Atom
Exercises | Q 24 | Page 385

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