Enegry associated with an electron of hydrogen atom is given by the expression

Options

A. `R_(H)((1)/(n^(2)))`
B. `-R_(H)n^(2)`
C. `R_(H)n^(2)`
D. `-R_(H)((1)/(n^(2)))`

Answer

D

Explanation

For `H` atom, `E_(n) = (-2pi^(2)me^(4))/(n^(2)h^(2)) =- (2pi^(2)me^(4))/(h^(2)) ((1)/(n^(2)))`
`E_(n) =- R_(H) ((1)/(n^(2))), n = 1, 2, 3,……`
where `R_(H)`, the proportionality constant is calleed Rydberg's constant and its value is `2.18 xx 10^(-18)J` or `13.6 eV`.