The velocity of an e in excited state of H-atom is `1.093 xx 10^(6)m//s`, what is the circumference of this orbit?

Options

A. `3.32 xx 10^(-10) m`
B. `6.64 xx 10^(-10) m`
C. `13.32 xx 10^(-10)m`
D. `13.28 xx 10^(-8)m`

Answer

C

Explanation

`V_(n) = 2.186 xx 10^(6)`
`rArr 1.093 xx 10^(6) = 2.186xx 10^(6) xx (1)/(n) , n = 2` from Bohr theory we know `2pi r = n lambda`
` = 2 lambda`, where `lambda = (h)/(mv)`
or `r = 0.529 (n^(2))/(Z) rArr 0.529 xx 4 Å`
`therefore` Circumference of the orbit .
`rArr 2 xx 0.529 xx 4 xx 10^(-10)`
`rArr 13.30 xx 10^(-10) m`