The radius of the orbit of an electron in a Hydrogen - like atom is `4.5s_(0)` where `s_(0)` is the bohr radius its orbital angular momentum is `(3b)/(2 pi) ` it is given that is is plank constant and R is rabdery constant .The possible wavelength `(s)` , when the atom de- exciter , is (are)

Options

A. `(9)/(32R)`
B. `(9)/(16R)`
C. `(9)/(5R)`
D. `(4)/(3R)`

Answer

A, C

Explanation

angular momentum ` = (nh)/(2 pi) = (3h)/(2 pi) therefore n = 3 `
`also r_(n) = (a_(0) n^(2))/(z) = 4.5 a_(0)`
:. (n^(2))/(z) = 4.5 rArr (9)/(z) = 4.5 rArr z = 2 `
we know that
`(1)/(lambda) = R z^(2) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] = (1)/(lambda) 4R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))] `
`for n_(2) = 3 , n_(1) = 1 ` we get `lambda= (9)/(8 xx 48) = (9)/(32R) `
`for n_(2) = 3 , n_(1) = 2 ` we get `lambda = (36)/(5 xx 48) = (9)/(5R) `
``for n_(2) = 2 , n_(1) = 1 ` we get `lambda= (4)/(3 xx 48) = (1)/(3R)`
(a),(c ) are currect option