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Question
Find the frequency of revolution of an electron in Bohr’s 2nd orbit; if the radius and speed of electron in that orbit is 2.14 × 10-10 m and 1.09 × 106 m/s respectively. [π= 3.142]
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Solution 1
Given
r2 = 2.14 × 10-10 m
n = 2
v2 = 1.09 × 106 m/s
To find: Frequency of revolution (`nu_2`)
`v = romega=r(2pinu)`
`nu=v/(2pir)`
`nu_2=v_2/(2pir_2)=(1.09xx10^6)/(2xx3.142xx2.14xx10^-10)`
`nu_2=8.11xx10^14 Hz`
The frequency of revolution of electron in 2nd Bohr orbit is 8.11 × 1014 Hz.
Solution 2
`T = (2pir)/V`
`because T=1/f`
`f = V/(2pir)`
`f=(1.09 xx 10^6)/(2 xx 3.14 xx 2.14 xx 10^-10)`
`f=8.11 xx 10^14 Hz`
The frequency of revolution of electron in 2nd Bohr orbit is 8.11 × 1014 Hz.
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