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Question
Find the value of energy of electron in eV in the third Bohr orbit of hydrogen atom.
(Rydberg's constant (R) = 1· 097 x 107m - 1,Planck's constant (h) =6·63x10-34 J-s,Velocity of light in air (c) = 3 x 108m/ s.)
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Solution
R = 1.097 × 107 m-1
h = 6.63 × 10-34 J - s ,
c = 3 × 108 m/s
n = 3
e = 1.6 × 10-19 C ,
E3 = ?
E = `(-6.63 xx 10^-34 xx 1.097 xx 10^7 xx 3 xx 10^8)/(9 xx 1.6 xx 10^-19)`
`= (-6.63 xx 1.097 xx 3)/(9 xx 1.6)`
`= - ["log" (6.63) + "log" (1.097) + "log"(3) - "log"(9) - "log" (1.6)]`
`[0.8215 + 0.0402 + 0.4771 - 0.9542 - 0.2041]`
`["antilog" (0.1805)]`
= 1.515 eV
The value of energy of electron in the third bohr orbit of hydrogen atom is 1.515 eV .
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