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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.
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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 × 105 revolution
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