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Energy of an electron in an excited hydrogen atom is `-3.4eV`. Its angualr momentum will be: `h = 6.626 xx 10^(-34) J-s`.
A. `2.11 xx 10^(-34)`
B. `3 xx 10^(-34)`
C. `1.055 xx 10^(-34)`
D. `0.5 xx 10^(-34)`

1 Answer

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Best answer
Correct Answer - A
`E_(n) = - 3.4 e V, E_(n) prop - (1)/(n^(2))`
` E_(1) = - 13.6 e V`.
Clearly `n = 2`
`= (n h)/(2 pi) = (2 h)/(2 pi) = (h)/(pi) = 2.11 xx 10^(-34) J s`

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