If the radius of electron orbit in the excited state of hydrogen atom is 476.1 pm, the energy of electron in that excited state in J is Radius and energy of electron in the first orbit of hydrogen atom are 52.9 pm and `-218xx10^(-18) J` respectively)

Options

A. `-2.42xx10^(-18)`
B. `19.62xx10^(-18)`
C. `-2.42xx10^(19)`
D. `-6.05xx10^(-19)`

Answer

C

Explanation

Given : Radius of excited state of hydrogen atom `=476 cdot 1 PM`
`E_(1) =2.18xx10^(-8) J"/ atom"`
`E_(n) =-2.18xx10^(-18) cdot (Z^(2))/(n^(2))`
`r_(1) =52.9 pm`
`r_(n) =(n^(2)xx527)/(Z), 476 cdot 1=(n^(2)xx52.7)/(Z)` [Z for hydrogen =1]
`therefore n^(2)=9 and E_(n) =E_(3) =-(2.18Z^(2)xx10^(-18)J)/(9)`
`=-2.42xx10^(-19) J`