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Questions
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?
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?
[Given: e = 1.6 × 10-19 C, m = 9.1 × 10-31 kg]
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Solution
Bohr Model Background
The time period Tn for an electron in the nnn-th orbit is given by:
`T_n = (2pir_n)/v_n`
Radius of n-th orbit:
rn = n2 ⋅ r1 where r1 = 5.29 × 10−11 m
Speed in n-th orbit:
`v_n = v_1/n "where" v_1 = 2.18 xx 10^6` m/s
So for n = 2:
r2 = 4 ⋅ 5.29 × 10−11 = 2.116 × 10−10 m
`v^2 = (2.18xx10^6)/2 = 1.09 xx 10^6` m/s
`T_2 = (2pir_2)/v_2 = (2pixx2.116 xx 10^-10)/(1.09 xx 10^6)`
`T_2 = (13.29 xx 10^-10)/(1.09 xx 10^6)` = 1.219×10−15s
t = 10−8s
Number of revolutions `= t/T^2 = 10^-8/1.219xx10^-15`
= 8.2 × 106 revolutions
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