The radius of orbit of an electron and the speed of electron in the ground state of hydrogen atom are ` 5 . 5 xx 10 ^(-11) m and 4 xx 10^(6) ms^(-1)` respectivley . Then , the orbital period of this electron in the first excited state will be . . . . . .

Options

A. `6 . 908 xx 10^(-6) s`
B. `9 . 608 xx 10^(-16) s`
C. `7 . 806 xx 10^(-16) s`
D. `8 . 9068 xx 10 ^(-16) s`

Answer

A

Explanation

Given in ground state the radius and velocity of electron in hydrogen atom
`r_(1) = 5 . 5 xx 10^(-11) m `
`v_(1) = 4 xx 10^(6) m//s`
In first excited state
`r_(n) prop (n^(2))/(Z)`
`r_(n) = 5 . 5 xx 10^(-11) xx 2^(2)`
The speed of electron
`v_(n) prop (Z)/(n)`
`= (4 xx 10^(6))/(2)`
`= 2 xx 10^(6) m//s`
The time period of revolution
`T = (2pir)/(v_(n))`
`= (2 xx 3.14 xx 220 xx 10^(-11))/(2 xx 10^(6))`
`= 6 . 908 xx 10^(-16) s`